1.           Sketch the graphs of:

$\begin{array}{l} \text{(a)}\ \ \ y=x-1\\\\ \text{(b)}\ \ \ y=-x-2\\\\ \text{(c)}\ \ \ y=-x+2\\\\ \text{(d)}\ \ \ y=2 x+1\\\\ \text{(e)}\ \ \ y=3 x^{2}\\\\ \text{(f)}\ \ \ y=-3 x^{2}\\\\ \text{(g)}\ \ \ y=\frac{1}{3} x^{2}\\\\ \text{(h)}\ \ \ y=\sqrt{2 x}\\\\ \text{(a)}\ \ \ y=|2 x|\\\\ \end{array}$

(a)       $y=x-1$

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Firstly, we have to construct a table of order pair $(x,y)$ for the graph $y=x-1$ and plotting them on the cartesian plane. Then join the points with smooth curve to illustrate the required graph.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=x-1 & \ldots & -4 & -3 & -2 & -1 & 0 & 1 & 2 & \ldots \\ \hline \end{array}$

(b)       $y=-x-2$

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Firstly, we have to construct a table for order pair $(x,y)$ for the graph $y=-x-2$ and plotting them on the cartesian plane. Then join the points with smooth curve to illustrate the required graph.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=-x-2 & \ldots & 1 & 0 & -1 & -2 & -3 & -4 & -5 & \ldots \\ \hline \end{array}$

(c)       $y=-x+2$

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Firstly, we have to construct a table for order pair $(x,y)$ for the graph $y=-x+2$ and plotting them on the cartesian plane. Then join the points with smooth curve to illustrate the required graph.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=-x+2 & \ldots & 5 & 4 & 3 & 2 & 1 & 0 & -1 & \ldots \\ \hline \end{array}$

(d)       $y=2x+1$

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Firstly, we have to construct a table for order pair $(x,y)$ for the graph $y=2x+1$ and plotting them on the cartesian plane. Then join the points with smooth curve to illustrate the required graph.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=2x+1 & \ldots & -5 & -3 & -1 & 1 & 3 & 5 & 7 & \ldots \\ \hline \end{array}$

(e)       $y=3x^2$

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Firstly, we have to construct a table for order pair $(x,y)$ for the graph $y=3x^2$ and plotting them on the cartesian plane. Then join the points with smooth curve to illustrate the required graph.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=3x^2 & \ldots & 27 & 12 & 3 & 0 & 3 & 12 & 27 & \ldots \\ \hline \end{array}$

(f)       $y=-3x^2$

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Firstly, we have to construct a table for order pair $(x,y)$ for the graph $y=-3x^2$ and plotting them on the cartesian plane. Then join the points with smooth curve to illustrate the required graph.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=-3x^2 & \ldots & -27 & -12 & -3 & 0 & -3 & -12 & -27 & \ldots \\ \hline \end{array}$

(g)       $y=\displaystyle\frac{1}{3}x^2$

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Firstly, we have to construct a table for order pair $(x,y)$ for the graph $y=\displaystyle\frac{1}{3}x^2$ and plotting them on the cartesian plane. Then join the points with smooth curve to illustrate the required graph.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=\displaystyle\frac{1}{3}x^2 & \ldots & 3 & \displaystyle\frac{4}{3} & \displaystyle\frac{1}{3} & 0 & \displaystyle\frac{1}{3} & \displaystyle\frac{4}{3} & 3 & \ldots \\ \hline \end{array}$

(h)       $y=\sqrt{2x}$

Show/Hide Solution

Firstly, we have to construct a table for order pair $(x,y)$ for the graph $y=\sqrt{2x}$ and plotting them on the cartesian plane. Then join the points with smooth curve to illustrate the required graph.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & 0 & 1 & 2 & 3 & 4 & 6 & 6 & \ldots \\ \hline y=y=\sqrt{2x} & \ldots & 0 & \sqrt{2} & 2 & \sqrt{6} & 2\sqrt{2} & \sqrt{10} & 2\sqrt{3} & \ldots \\ \hline \end{array}$

(i)       $y=|2x|$

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Firstly, we have to construct a table for order pair $(x,y)$ for the graph $y=|2x|$ and plotting them on the cartesian plane. Then join the points with smooth curve to illustrate the required graph.

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=|2x| & \ldots & 6 & 4 & 2 & 0 & 2 & 4 & 6 & \ldots \\ \hline \end{array}$

Grade 10 ဟာ function ရဲ့ အခြေခံသဘောကို သင်ကြားတာ ဖြစ်ပါတယ်။ ဒါကြောင့် ကျောင်းသူကျောင်းသားများ အနေဖြင့် function တစ်ခုကို graph ဆွဲတဲ့အခါ graph paper ကို သုံးရပါမယ်။ graph paper မသုံးပဲဆွဲလို့မရဘူးဆိုတော့ ရပါတယ်။ graph တစ်ခုတည်းမဟုတ်ပဲ graph များစွာဆွဲရန် လိုအပ်တဲ့အခါ graph paper သုံးပြီး ဆွဲမှသာ အဆင်ပြေပါတယ်။

သဘောတရားပဲ ဆွဲပြရင်မရဘူးလားလို့ စောဒက တက်စရာရှိပါတယ်။ function တစ်ခုရဲ့ သဘောတရားနဲ့ သဘောသဘာ၀ကို သိဖို့ ကျောင်းသားတစ်ယောက်နေနဲ့ $(x, y)$ order pair တွေကို များနိုင်သမျှ များများရှာထားမှသာ အဆင်ပြေပါတယ်။ ယေဘုယျ ပြောရရင်တော့ grade 10 အဆင့် သင်ခန်းစာတွေမှာ domain က real number ဖြစ်တဲ့အခါ integer subset တစ်ခု ဖြစ်တဲ့ $\{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5 \}$ ထည့်ဆွဲမယ်ဆိုရင် function ရဲ့ nature ကို ပေါ်လွင်စေနိုင်ပါတယ်။...

2.        Sketch the graphs of $y =2x$ and $y =\displaystyle\frac{1}{2}x$ in the same plane. What do you notice from the graphs? Explain.

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$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=2x & \ldots & -6 & -4 & -2 & 0 & 2 & 4 & 6 & \ldots \\ \hline \end{array}$

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=\displaystyle \frac{1}{2}x & \ldots & -\displaystyle \frac{3}{2} & -1 & -\displaystyle \frac{1}{2} & 0 & \displaystyle \frac{1}{2} & 1 & \displaystyle \frac{3}{2} & \ldots \\ \hline \end{array}$

• Both are linear functions.

• Both graphs are straight lines.

• Both are increasing, i.e., both have positive slopes.

• Both passes through the origin.

3.        Sketch the graphs of $y =\displaystyle \frac{1}{2}x^2$ and $y =2x^2$ in the same plane. What do you notice from the graphs? Explain.

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$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=\displaystyle \frac{1}{2}x^2 & \ldots & \displaystyle \frac{9}{2} & 2 & \displaystyle \frac{1}{2} & 0 & \displaystyle \frac{1}{2} & 2 & \displaystyle \frac{9}{2} & \ldots \\ \hline \end{array}$

$\begin{array}{|c|c|c|c|c|c|c|c|c|c|} \hline x & \ldots & -3 & -2 & -1 & 0 & 1 & 2 & 3 & \ldots \\ \hline y=2x^2 & \ldots & 18 & 8 & 2 & 0 & 2 & 8 & 18 & \ldots \\ \hline \end{array}$

• Both graphs are parabolas.

• Both are opening upward.

• Both functions have vertices at the origin.

#### How to Determine the Domain of a Function

When the domain of a function is not specified, then assume that it is the set of all possible real numbers for which the function makes sense.

ပေးထားသော function တစ်ခုအတွက် domain သတ်မှတ်ပေးထားခြင်း မရှိလျှင် function ၏ သတ်မှတ်ချက်ကို ပြေလည်စေသော အစုဝင်အားလုံးပါသည် ကိန်းစစ်အစုသည် အဆိုပါ function ၏ domain ဖြစ်သည်ဟု မှတ်ယူရမည်။

#### Equality of Functions

Two functions $f$ and $g$ are equal (and we write $f=g$) if and only if

1. $f$ and $g$ have the same domain, and

2. $f(c)$ = $g(c)$ for each element $c$ in the domain.

Function နှစ်ခု $f$ နှင့် $g$ ၏ domain များတူညီကြပြီး domain ထဲရှိ အစုဝင် $c$ တိုင်းအတွက် $f(c)= g(c)$ [$c$ ၏ image တူညီလျှင်] ဖြစ်လျှင် $f$ နှင့် $g$ သည် တူညီသော function များ ဖြစ်ကြသည်။

1.           Determine whether each relation is a function or not. If it is a function, state the domain and range.

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$\begin{array}{ll} \text{(a)} & \text{The relation}\ R\ \text{is a function.}\\\\ & \text{dom}\ (R ) = \{-1, 0, 1\}\\\\ & \text{ran}\ (R) = \{0, 1\}\\\\ \text{(b)} & \text{The relation}\ R\ \text{is not a function.}\\\\ \text{(c)} & \text{The relation}\ R\ \text{is not a function.} \end{array}$

2.           Consider the following relations. Determine whether each relation is a function or not. If it is a function, write down the domain and range.

$\begin{array}{l} \text{(a)}\ \ \{(1,3),(2,5),(3,7),(4,9)\} \\\\ \text{(b)}\ \ \{(-2,5),(-1,3),(0,1),(-1,1)\}\\\\ \text{(c)}\ \ \{(1,3),(2,3),(3,2),(2,1)\} \\\\ \text{(d)}\ \ \{(2,4),(3,6),(4,6),(7,14)\} \\\\ \text{(e)}\ \ \{(0,0),(1,1),(3,3),(4,4)\} \\\\ \text{(f)}\ \ \{(2, a),(4, c),(5, a),(4, e)\} \end{array}$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\;\;\{(1,3),(2,5),(3,7),(4,9)\}\\\ \ \ \ \ \text{The given relation is a function because }\\\ \ \ \ \ \text{each element of domain is related}\\\ \ \ \ \ \text{to exactly one element}\text{.}\\\ \ \ \ \ \text{Domain}=\{1,\ 2,\ 3,\ 4\}\\\ \ \ \ \ \text{Range}=\{3,\ 5,\ 7,\ 9\}\\\\\text{(b)}\;\;\{(-2,5),(-1,3),(0,1),(-1,1)\}\\\ \ \ \ \ \text{The given relation is not}\ \text{a function}\ \text{because }\\\ \ \ \ \ \text{the element }-\text{1 of domain is related to two}\\\ \ \ \ \ \text{elements 3 and 1}\text{.}\\\\\text{(c)}\;\;\{(1,3),(2,3),(3,2),(2,1)\}\\\ \ \ \ \ \text{The given relation is not}\ \text{a function}\ \text{because }\\\ \ \ \ \ \text{the element }2\text{ of domain is related to two}\\\ \ \ \ \ \text{elements 3 and 1}\text{.}\\\\\text{(d)}\;\;\{(2,4),(3,6),(4,6),(7,14)\}\\\ \ \ \ \ \text{The given relation is a function because }\\\ \ \ \ \ \text{each element of domain is related}\\\ \ \ \ \ \text{to exactly one element}\text{.}\\\ \ \ \ \ \text{Domain}=\{2,\ 3,\ 4,\ 7\}\\\ \ \ \ \ \text{Range}=\{4,\ 6,\ 14\}\\\\\text{(e)}\;\;\{(0,0),(1,1),(3,3),(4,4)\}\\\ \ \ \ \ \text{The given relation is a function because }\\\ \ \ \ \ \text{each element of domain is related}\\\ \ \ \ \ \text{to exactly one element}\text{.}\\\ \ \ \ \ \text{Domain}=\{0,\ 1,\ 3,\ 4\}\\\ \ \ \ \ \text{Range}=\{0,\ 1,\ 3,\ 4\}\\\\\text{(f)}\;\;\{(2,a),(4,c),(5,a),(4,e)\}\\\ \ \ \ \ \text{The given relation is not}\ \text{a function}\ \text{because }\\\ \ \ \ \ \text{the element }4\text{ of domain is related to two}\\\ \ \ \ \ \text{elements }c\text{ and }e\text{.}\end{array}$

3.           Let $f$ be a function from $\mathbb{R} \rightarrow \mathbb{R}$. Which of the following statements are true?

(a)     If $f(x)=5-x,$ the image of -3 under $f$ is 8.

(b)     If $f(x)=x^{2}+9,$ the image of -3 under $f$ is zero.

(c)     If $f(x)=3 x+4,$ then $f(a)=a$ implies that $a=-2$.

(d)     If $f(x)=x+3,$ there is only one value $a \in \mathbb{R}$ such that $f(a)=0$.

(e)     If $f(x)=x^{2}-1,$ then there are exactly two values $a \in \mathbb{R}$ such that $f(a)=0$.

Show/Hide Solution

$\begin{array}{l}\ \ \ \ \ \ \ \ f:\mathbb{R}\to \mathbb{R}\\\\\text{(a)}\ \ f(x)=5-x\\\ \ \ \ \ f(-3)=5-(-3)=8\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\\\\\text{(b)}\ \ f(x)={{x}^{2}}+9\\\ \ \ \ \ f(-3)={{(-3)}^{2}}+9=18\\\ \ \ \ \ \therefore \ \text{ The statements is false}\text{.}\\\\\text{(c)}\ \ f(x)=3x+4\\\ \ \ \ \ f(a)=a\\\ \ \ \ \ 3a+4=a\\\ \ \ \ \ 2a=-4\\\ \ \ \ \ a=-2\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\\\\\text{(d)}\ \ f(x)=x+3\\\ \ \ \ \ f(a)=0\\\ \ \ \ \ a+3=0\\\ \ \ \ \ a=-3\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\\\\\text{(e)}\ \ f(x)={{x}^{2}}-1\\\ \ \ \ \ f(a)=0\\\ \ \ \ \ {{a}^{2}}-1=0\\\ \ \ \ \ {{a}^{2}}=1\\\ \ \ \ \ a=\pm 1\\\ \ \ \ \ \therefore \ \text{ The statements is true}\text{.}\end{array}$

4.           Illustrate the function $f: x \mapsto x+2$ with an arrow diagram for the domain $\{3,5,7,9,10\} .$ Write down the range of $f$.

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$\begin{array}{l}\ \ \ f:x\mapsto x+2\\\\\ \ \ \therefore \ f(x)=x+2\\\\\ \ \ \operatorname{dom}(f)=\{3,5,7,9,10\}\\\\\ \ \ f(3)=3+2=5\\\\\ \ \ f(5)=5+2=7\\\\\ \ \ f(7)=7+2=9\\\\\ \ \ f(9)=9+2=11\\\\\ \ \ f(10)=10+2=12\\\\\ \ \ \operatorname{ran}(f)=\{5,7,9,11,12\}\end{array}$

5.           Let the domain of function $h: x \mapsto 0$ be $\{2,4,6,7\} .$ What is the range of $h ?$ Draw an arrow diagram for $h$.

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$\begin{array}{l}h:x\mapsto 0\\\\\operatorname{dom}(h)=\{2,4,6,7\}\\\\\therefore \ \ h:2\mapsto 0\\\\\ \ \ h:4\mapsto 0\\\\\ \ \ h:6\mapsto 0\\\\\ \ \ h:7\mapsto 0\\\\\operatorname{ran}(h)=\{0\}\end{array}$

6.           Let the domain of function $f: x \mapsto 3 x$ be the set of natural numbers less than $5 .$ State the domain and range.

Show/Hide Solution

$\begin{array}{l} f:x\mapsto 3x\\\\ \operatorname{dom}(f)=\text{the set of natural numbers less}\ \text{than 5}\text{.}\\\\ \operatorname{dom}(f)=\{1,\ 2,\ 3,\ 4\}\\\\ f:1\mapsto 3(1)=3\\f:2\mapsto 3(2)=6\\\\ f:3\mapsto 3(3)=9\\f:4\mapsto 3(4)=12\\\\ \operatorname{ran}(f)=\{3,\ 6,\ 9,\ 12\}\ end{array}$

7.           Let $g: \mathbb{R} \rightarrow \mathbb{R}$ be given by

(a)     $g(x)=3-4 x$ .Find $g(1), \quad g(3), \quad g(-2), \quad g(x+3), \quad g\left(\displaystyle \frac{1}{2}\right).$

(b)    $g(x)=2 x-5 .$ Find $g(3), g\left(\displaystyle\frac{1}{2}\right), g(0), g(-4), g(4) .$ If $g(a)=99,$ find $a$.

(c)    $g(x)=\displaystyle\frac{x+5}{2} .$ Find the images of $3,0,-3 .$ Find $x$ if $g(x)=0$.

(d)    $g(x)=3 x-1 .$ Find $x$ such that $g(x)=20$.

Show/Hide Solution

$\begin{array}{ll} \text{(a)} & g(x) = 3-4x\\ & g(1) = 3-4(1) = -1\\ & g(3) = 3-4(3) = -9\\ & g(-2) = 3-4(-2) = 11\\ & g(x+3) = & 3-4(x+3) = -4x-9\\ & g \left({\displaystyle\frac{1}{2}} \right) = 3- 4\left(\displaystyle\frac{1}{2}\right) = 1\\\\ \text{(b)} & g(x) = 2x-5\\ & g(3) = 2(3)-5 = 1\\ & g \left(\displaystyle\frac{1}{2}\right) = 2 \left(\displaystyle\frac{1}{2}\right)-5 = -4\\ & g(0) = 2(0)-5 = -5\\ & g(-4) = 2(-4)-5 = -13\\ & g(4) = 2(4)-5 = 3\\\\ & g(x) = 0\\ & 2x-5 = 0\\ & x = \displaystyle\frac{5}{2}\\\\ \text{(c)} & g(x) = \displaystyle\frac{x+5}{2}\\ & g(3) = \displaystyle\frac{3+5}{2} = 4\\ & g(0) = \displaystyle\frac{0+5}{2} = \displaystyle\frac{5}{2}\\ & g(-3) = \displaystyle\frac{-3+5}{2} = 1\\\\ \text{(d)} & g(x) = 3x-1\\ & g(x) = 20\\ & 3x-1 = 20\\ & x = 7 \end{array}$

8.           A function $f$ from $A$ to $A$, where $A$ is the set of positive integers, is given by $f(x)=$ the sum of all possible divisors of $x$.

For example $f(6)=1+2+3+6=12$

(a)     Find the values of $f(2), f(5), f(13), f(18)$.

(b)     Show that $f(14)=f(15)$ and $f(3) \cdot f(5)=f(15)$.

Show/Hide Solution

$\begin{array}{l}f(x)=\text{the sum of all possible divisors of }x\\\\f(2)=1+2=3\\\\f(5)=1+5=6\\\\f(13)=1+13=14\\\\f(15)=1+3+5+15=24\\\\f(14)=1+2+7+14=24\\\\\therefore \ f(15)=f(14)\\\\f(3)=1+3=4\\\\f(3)\cdot f(5)=4\times 6=24\\\\\therefore \ f(3)\cdot f(5)=f(15)\end{array}$

9.           Let $A=$ the set of positive integers greater than 3 and $B=$ the set of all positive integers. Let $d: A \rightarrow B$ be a function given by $d(n)=\frac{1}{2} n(n-3)$ the number of diagonals of a polygon of $n$ sides.

(a)     Find $d(6), d(8), d(10), d(12)$.

(b)     How many diagonals will a polygon of 20 sides have?

Show/Hide Solution

$\begin{array}{l}\text{(a)}\ \ \ \ A=\text{ the set of positive integers greater than 3}\\\\\ \ \ \ \ \ \ A=\{4,\ 5,\ 6,\ ...\}\\\\\ \ \ \ \ \ \ B=\text{ the set of positive all integers }\\\\\ \ \ \ \ \ \ B=\{1,\ 2,\ 3,\ ...\}\\\\\ \ \ \ \ \ \ d:A\to B\ \\\\\ \ \ \ \ \ d(n)=\displaystyle \frac{1}{2}n(n-3)\\\\\ \ \ \ \ \ d(6)=\displaystyle \frac{1}{2}\cdot 6\cdot (6-3)=9\\\\\ \ \ \ \ \ d(8)=\displaystyle \frac{1}{2}\cdot 8\cdot (8-3)=20\\\\\ \ \ \ \ \ d(10)=\displaystyle \frac{1}{2}\cdot 10\cdot (10-3)=35\\\\\ \ \ \ \ \ d(12)=\displaystyle \frac{1}{2}\cdot 12\cdot (12-3)=54\\\\\text{(b)}\ \ \ d(20)=\displaystyle \frac{1}{2}\cdot 20\cdot (20-3)=170\end{array}$

10.           Determine whether $f$ and $g$ are equal functions or not. Give reason:

(a)     $f(x)=x^{2}+2,\ g(x)=(x+2)^{2}$.

(b)     $f(x)=x^{2},\ g(x)=|x|^{2}$.

(c)     $f(x)=\displaystyle\frac{x^{2}-1}{x+1},\ g(x)=x-1$.

(d)     $f(x)=\displaystyle\frac{x+2}{x^{2}-4},\ g(x)=\frac{1}{x-2}$.

Show/Hide Solution

$\begin{array}{l} \text{(a)}\ \ f(x)={{x}^{2}}+2,\;g(x)={{(x+2)}^{2}}\\ \ \ \ \ \ \operatorname{dom}(f)=\mathbb{R},\ \operatorname{dom}(g)=\mathbb{R}\\ \,\,\,\ \ \ f(1)={{1}^{2}}+2=3\\\ \ \ \ \ g(1)={{(1+2)}^{2}}=9\\ \ \ \ \ \ \therefore \ \operatorname{ran}(f)\ne \operatorname{ran}(g)\\ \ \ \ \ \ \therefore \ f\ne g\\\\ \text{(b)}\ \ f(x)={{x}^{2}},\;g(x)=|x{{|}^{2}}\\ \ \ \ \ \ \operatorname{dom}(f)=\mathbb{R},\ \operatorname{dom}(g)=\mathbb{R}\\ \,\,\,\ \ \ \operatorname{ran}(f)=\{x\ |\ x\ge 0,\ x\in \mathbb{R}\}\\ \,\,\,\ \ \ \operatorname{ran}(g)=\{x\ |\ x\ge 0,\ x\in \mathbb{R}\}\\ \ \ \ \ \ \therefore \ \operatorname{dom}(f)=\operatorname{dom}(g)\ \ \text{and }\operatorname{ran}(f)=\operatorname{ran}(g)\\ \ \ \ \ \ \therefore \ f=g\\\\ \text{(c)}\ \ f(x)=\displaystyle \frac{{{{x}^{2}}-1}}{{x+1}},\;g(x)=x-1\\ \ \ \ \ f(x)\ \text{exists only when }x+1\ne 0,\ \text{i}\text{.e,}\ x\ne -1\\ \ \ \ \ \ \operatorname{dom}(f)=\mathbb{R}\smallsetminus \{-1\},\ \operatorname{dom}(g)=\mathbb{R}\\ \ \ \ \ \ \therefore \ \operatorname{dom}(f)\ne \operatorname{dom}(g)\\ \ \ \ \ \ \therefore \ f\ne g\\\\ \text{(d)}\ \ f(x)=\displaystyle \frac{{x+2}}{{{{x}^{2}}-4}},\;g(x)=\displaystyle \frac{1}{{x-2}}\\ \ \ \ \ \ f(x)\ \text{exists only when }\\ \,\ \ \ \ {{x}^{2}}-4\ne 0\\ \ \ \ \ \ {{x}^{2}}\ne 4\\ \ \ \ \ \ \therefore x\ne \pm \ 2\\ \ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\mathbb{R}\smallsetminus \{\pm \ 2\}\\ \ \ \ \ \ g(x)\ \text{exists only when }\\ \,\ \ \ \ x-2\ne 0\\\ \ \ \ \ x\ne 2\\ \ \ \ \ \ \therefore \ \operatorname{dom}(g)=\mathbb{R}\smallsetminus \{2\}\\ \ \ \ \ \ \therefore \ \operatorname{dom}(f)\ne \operatorname{dom}(g)\\ \ \ \ \ \ \therefore \ f\ne g\end{array}$

11.           State the domain of the following functions.

(a)     $f(x)=\sqrt{x-2}$.

(b)     $f(x)=\displaystyle\frac{1}{2 x-1}$.

(c)     $f(x)=\displaystyle\frac{4}{x-3}$.

(d)     $f(x)=\displaystyle\frac{2}{x^{2}-1}$.

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$\begin{array}{l}\text{(a)}\ \ f(x)=\sqrt{{x-2}}\\\ \ \ \ \ f(x)\ \text{exists only when}\\\ \ \ \ \ x-2\ge 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ \ x\ge 2\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ge 2,\ x\in \mathbb{R}\}\\\\\text{(b)}\ \ f(x)=\displaystyle \frac{1}{{2x-1}}\\\ \ \ \ \ f(x)\ \text{exists only when}\\\ \ \ \ \ 2x-1\ne 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ \ x\ne \displaystyle \frac{1}{2}\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ne \displaystyle \frac{1}{2},\ x\in \mathbb{R}\}\\\\\text{(c)}\ \ f(x)=\displaystyle \frac{4}{{x-3}}\\\ \ \ \ \ f(x)\ \text{exists only when}\\\ \ \ \ \ x-3\ne 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ \ x\ne 3\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ne 3,\ x\in \mathbb{R}\}\\\\\text{(d)}\ \ f(x)=\displaystyle \frac{2}{{{{x}^{2}}-1}}\\\ \ \ \ \ f(x)\ \text{exists only when }\\\,\ \ \ \ {{x}^{2}}-1\ne 0\\\ \ \ \ \ \text{i}\text{.e}\text{.,}\ {{x}^{2}}\ne 1\\\ \ \ \ \ \therefore x\ne \pm \ 1\\\ \ \ \ \ \therefore \ \ \operatorname{dom}(f)=\{x\ |\ x\ne \pm \ 1,\ x\in \mathbb{R}\}\end{array}$

$\begin{array}{|l|l|l|} \hline {\ \ \ \ \ \ \ \ \ \ \text { Properties }} & {\ \ \ \ \ \ \ \text { For Exponents }} & {\ \ \ \ \ \ \ \ \ \ \ \ \ \text { For Logarithms }} \\ \hline \text { One-to-one Property } & \text { If } b^{x}=b^{y}, \text { then } x=y & \text { If } \log _{b} M=\log _{b} N, \text { then } M=N \\ \hline \text { Product Property } & b^{x} \cdot b^{y}=b^{x+y} & \log _{b}(M N)=\log _{b} M+\log _{b} N \\ \hline \text { Quotient Property } & \displaystyle\frac{b^{x}}{b^{y}}=b^{x-y} & \log _{b} \displaystyle\frac{M}{N}=\log _{b} M-\log _{b} N \\ \hline \text { Power Property } & \left(b^{x}\right)^{y}=b^{x y} & \log _{b} N^{p}=p \log _{b} N \\ \hline \end{array}$

1.           Replace $\square$ with the appropriate number.

$\begin{array}{l} \text{(a)}\ \ \log _{3} 24=\log _{3} 6+\log _{3} \square\\ \text{(b)}\ \ \log _{5} 24=\log _{5} 60+\log _{5} \square\\ \text{(c)}\ \ \log _{2} \square=3 \log _{2} 3\\ \text{(d)}\ \ \log _{10} 9=\square \log _{10} 3\\ \text{(e)}\ \ \log _{8} 5=\log _{8} \square-\log _{8} 11 \end{array}$

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$\begin{array}{l} \text{(a)}\ \ 4\\ \text{(b)}\ \ \displaystyle\frac{2}{5}\\ \text{(c)}\ \ 27\\ \text{(d)}\ \ 2\\ \text{(e)}\ \ 55 \end{array}$

2.           Write each expression as a single logarithm.

$\begin{array}{l} \text{(a)}\ \ \log _{b} 20+\log _{b} 57-\log _{b} 241\\ \text{(b)}\ \ 3 \log _{b} 8-\displaystyle\frac{1}{2} \log _{b} 12\\ \text{(c)}\ \ \log _{b} x-2 \log _{b} y-\log _{b} a\\ \text{(d)}\ \ \log _{2} 3+\log _{4} 15 \end{array}$

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$\begin{array}{l}\text{(a)}\;\;\ \ {{\log }_{b}}20+{{\log }_{b}}57-{{\log }_{b}}241\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{20\times 57}}{{241}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{1140}}{{241}}\\\\ \text{(b)}\;\;\ \ 3{{\log }_{b}}8-\displaystyle \frac{1}{2}{{\log }_{b}}12\\ \ \ \ \ =\ {{\log }_{b}}{{8}^{3}}-{{\log }_{b}}{{12}^{{\displaystyle \frac{1}{2}}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{8\times 8\times 8}}{{\sqrt{{12}}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{8\times 8\times 8}}{{2\sqrt{3}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{256}}{{\sqrt{3}}}\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{{256\sqrt{3}}}{3}\\\\ \text{(c)}\;\;\ \ {{\log }_{b}}x-2{{\log }_{b}}y-{{\log }_{b}}a\\ \ \ \ \ =\ {{\log }_{b}}x-{{\log }_{b}}{{y}^{2}}-{{\log }_{b}}a\\ \ \ \ \ =\ {{\log }_{b}}\displaystyle \frac{x}{{a{{y}^{2}}}}\\\\ \text{(d)}\;\;\ \ {{\log }_{2}}3+{{\log }_{4}}15\\ \ \ \ \ \text{Let}\ {{\log }_{4}}15=x,\ \text{then}\\ \ \ \ \ 15={{4}^{x}}\\\ \ \ \ 15={{2}^{2}}^{x}\\ \ \ \ \ \therefore \ \ 2x={{\log }_{2}}15\\ \ \ \ \ \therefore \ \ x=\displaystyle \frac{1}{2}{{\log }_{2}}15\\ \ \ \ \ \therefore \ \ {{\log }_{4}}15={{\log }_{2}}\sqrt{{15}}\\ \ \ \ \ \therefore \ \ {{\log }_{2}}3+{{\log }_{4}}15\\ \ \ \ \ =\ \ {{\log }_{2}}3+{{\log }_{2}}\sqrt{{15}}\\ \ \ \ \ =\ \ {{\log }_{2}}3\sqrt{{15}} \end{array}$

3.           Write each expression in terms of $\log _{b} 2, \log _{b} 3$ and $\log _{b} 5$.

$\begin{array}{l} \text{(a)}\ \ \log _{b} 8\\ \text{(b)}\ \ \log _{b} 15\\ \text{(c)}\ \ \log _{b} 270\\ \text{(d)}\ \ \log _{b} \displaystyle\frac{27 \sqrt[3]{5}}{16}\\ \text{(e)}\ \ \log _{b} \displaystyle\frac{216}{\sqrt[3]{32}}\\ \text{(f)}\ \ \log _{b}(648 \sqrt{125})\\ \end{array}$

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$\begin{array}{l} \text{(a)}\;\;{{\log }_{b}}8={{\log }_{b}}{{2}^{3}}=3{{\log }_{b}}2\\\\ \text{(b)}\;\;{{\log }_{b}}15={{\log }_{b}}(3\times 5)={{\log }_{b}}3+{{\log }_{b}}5\\\\ \text{(c)}\;\;{{\log }_{b}}270={{\log }_{b}}(2\times {{3}^{3}}\times 5)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}2+{{\log }_{b}}{{3}^{3}}+{{\log }_{b}}5\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}2+3{{\log }_{b}}3+{{\log }_{b}}5\\\\ \text{(d)}\;\;{{\log }_{b}}\displaystyle \frac{{27\sqrt[3]{5}}}{{16}}={{\log }_{b}}\displaystyle \frac{{{{3}^{3}}\times {{5}^{{\frac{1}{3}}}}}}{{{{2}^{4}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}{{3}^{3}}+{{\log }_{b}}{{5}^{{\frac{1}{3}}}}-{{\log }_{b}}{{2}^{4}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3{{\log }_{b}}3+\displaystyle \frac{1}{3}{{\log }_{b}}5-4{{\log }_{b}}2\\\\ \text{(e)}\;\;{{\log }_{b}}\displaystyle \frac{{216}}{{\sqrt[3]{{32}}}}={{\log }_{b}}\displaystyle \frac{{{{2}^{3}}\times {{3}^{3}}}}{{{{2}^{{\frac{5}{3}}}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}\left( {{{2}^{{^{{\frac{4}{3}}}}}}\times {{3}^{3}}} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\displaystyle \frac{4}{3}{{\log }_{b}}2+3{{\log }_{b}}3\\\\ \text{(f)}\;\;{{\log }_{b}}(648\sqrt{{125}})={{\log }_{b}}({{2}^{3}}\times {{3}^{4}}\times {{5}^{{\frac{3}{2}}}})\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{b}}{{2}^{3}}+{{\log }_{b}}{{3}^{4}}+{{\log }_{b}}{{5}^{{\frac{3}{2}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3{{\log }_{b}}2+4{{\log }_{b}}3+\displaystyle\frac{3}{2}{{\log }_{b}}5 \end{array}$

4.           Evaluate each expression.

$\begin{array}{l} \text{(a)}\ \ \log _{2} 128\\ \text{(b)}\ \ \log _{3} 81^{4}\\ \text{(c)}\ \ \log _{\frac{1}{2}} 8\\ \text{(d)}\ \ \log _{8} 2\\ \text{(e)}\ \ \log _{3} \displaystyle\frac{\sqrt{3}}{81}\\ \text{(f)}\ \ \displaystyle\frac{\log _{3} \sqrt{3}}{\log _{3} 81}\\ \text{(g)}\ \ \displaystyle\frac{\log _{2} 25}{\log _{2} 5}\\ \text{(h)}\ \ \log _{4} 8 \end{array}$

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$\begin{array}{l} \text{(a)}\;\;{{\log }_{2}}128={{\log }_{2}}{{2}^{7}}=7\\\\ \text{(b)}\;\;{{\log }_{3}}{{81}^{4}}={{\log }_{3}}{{\left( {{{3}^{4}}} \right)}^{4}}={{\log }_{3}}{{3}^{{16}}}=16\\\\ \text{(c)}\;\;{{\log }_{{\frac{1}{2}}}}8={{\log }_{{\frac{1}{2}}}}{{2}^{3}}={{\log }_{{\frac{1}{2}}}}{{\left( {\displaystyle\frac{1}{2}} \right)}^{{-3}}}=-3\\\\ \text{(d)}\;\;{{\log }_{8}}2={{\log }_{8}}{{8}^{{\frac{1}{3}}}}=\displaystyle \frac{1}{3}\\\\ \text{(e)}\;\;{{\log }_{3}}\displaystyle \frac{{\sqrt{3}}}{{81}}={{\log }_{3}}\displaystyle \frac{{{{3}^{{\frac{1}{2}}}}}}{{{{3}^{4}}}}={{\log }_{3}}{{3}^{{-\frac{7}{2}}}}=-\displaystyle \frac{7}{2}\\\\ \text{(f)}\;\;\displaystyle \frac{{{{{\log }}_{3}}\sqrt{3}}}{{{{{\log }}_{3}}81}}=\displaystyle \frac{{{{{\log }}_{3}}{{3}^{{\frac{1}{2}}}}}}{{{{{\log }}_{3}}{{3}^{4}}}}=\displaystyle \frac{{\frac{1}{2}}}{4}=\displaystyle \frac{1}{8}\\\\ \text{(g)}\;\;\displaystyle \frac{{{{{\log }}_{2}}25}}{{{{{\log }}_{2}}5}}=\displaystyle \frac{{{{{\log }}_{2}}{{5}^{2}}}}{{{{{\log }}_{2}}5}}=\displaystyle \frac{{2{{{\log }}_{2}}5}}{{{{{\log }}_{2}}5}}=2\\\\ \text{(h)}\;\;{{\log }_{4}}8={{\log }_{4}}\sqrt{{64}}={{\log }_{4}}{{4}^{{\frac{3}{2}}}}=\displaystyle \frac{3}{2} \end{array}$

5.           Use $\log _{10} 2=0.3010$ and $\log _{10} 3=0.4771$ to evaluate each of the following expressions.

$\begin{array}{lll} \text{(a)}\ \ \log _{10} 6 & \text{(b)}\ \ \log _{10} 1.5 & \text{(c)}\ \ \log _{10} \sqrt{3}\\\\ \text{(d)}\ \ \log _{10} 4 & \text{(e)}\ \ \log _{10} 4.5 & \text{(f)}\ \ \log _{10} 8\\\\ \text{(g)}\ \ \log _{10} 18 & \text{(h)}\ \ \log _{10} 5 \end{array}$

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$\begin{array}{l} {{\log }_{{10}}}2=0.3010,\ {{\log }_{{10}}}3=0.4771\\\\ \text{(a)}\;\;{{\log }_{{10}}}6=\;{{\log }_{{10}}}\left( {2\times 3} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}2+{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.3010+0.4771\\\ \ \ \ \ \ \ \ \ \ \ \ =0.7781\\\\ \text{(b)}\;\;{{\log }_{{10}}}1.5=\;{{\log }_{{10}}}\displaystyle \frac{3}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}3-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.4771-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.1761\\\\ \text{(c)}\;\;{{\log }_{{10}}}\sqrt{3}\ =\;{{\log }_{{10}}}{{3}^{{\frac{1}{2}}}}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;\displaystyle \frac{1}{2}{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;\displaystyle \frac{1}{2}\times 0.4771\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;0.2386\\\\\text{(d)}\;\;{{\log }_{{10}}}4={{\log }_{{10}}}{{2}^{2}}\\ \ \ \ \ \ \ \ \ \ \ \ \ =2{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =2\ (0.3010)\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.6020\\\\ \text{(e)}\;\;{{\log }_{{10}}}4.5=\;{{\log }_{{10}}}\displaystyle \frac{9}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}\displaystyle \frac{{{{3}^{2}}}}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;2{{\log }_{{10}}}3-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;2\left( {0.4771} \right)-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;0.6532\\\\ \text{(f)}\;\;{{\log }_{{10}}}8\ ={{\log }_{{10}}}{{2}^{3}}\\ \ \ \ \ \ \ \ \ \ \ \ \ =3{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =3\left( {0.3010} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.9030\\\\ \text{(g)}\;\;{{\log }_{{10}}}18={{\log }_{{10}}}\left( {2\times {{3}^{2}}} \right)\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =\;{{\log }_{{10}}}2+2{{\log }_{{10}}}3\\ \ \ \ \ \ \ \ \ \ \ \ \ \ =0.3010+2\left( {0.4771} \right)\\\\ \text{(h)}\;\;{{\log }_{{10}}}5={{\log }_{{10}}}\displaystyle \frac{{10}}{2}\\ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}10-{{\log }_{{10}}}2\\ \ \ \ \ \ \ \ \ \ \ \ \ =1-0.3010\\ \ \ \ \ \ \ \ \ \ \ \ \ =0.6990 \end{array}$

6.           Solve the following equations for $x$.

$\begin{array}{lll} \text{(a)}\ \ \log _{a} \displaystyle\frac{18}{5}+\log _{a} \displaystyle\frac{10}{3}-\log _{a} \displaystyle\frac{6}{7}=\log _{a} x\\\\ \text{(b)}\ \ \log _{b} x=2-a+\log _{b}\left(\displaystyle\frac{a^{2} b^{a}}{b^{2}}\right)\\\\ \text{(c)}\ \ \log x^{3}-\log x^{2}=\log 5 x-\log 4 x\\\\ \text{(d)}\ \ \log _{10} x+\log _{10} 3=\log _{10} 6\\\\ \text{(e)}\ \ 8 \log x=\log a^{\frac{3}{2}}+\log 2-\displaystyle\frac{1}{2} \log a^{3}-\log \frac{2}{a^{4}}\\\\ \end{array}$

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$\begin{array}{l} \text{(a)}\;\;{{\log }_{a}}\displaystyle \frac{{18}}{5}+{{\log }_{a}}\displaystyle \frac{{10}}{3}-{{\log }_{a}}\displaystyle \frac{6}{7}={{\log }_{a}}x\\ \ \ \ \ {{\log }_{a}}\left( {\displaystyle \frac{{\displaystyle \frac{{18}}{5}\times \displaystyle \frac{{10}}{3}}}{{\displaystyle \frac{6}{7}}}} \right)={{\log }_{a}}x\\ \ \ \ \ x=\displaystyle \frac{{18}}{5}\times \displaystyle \frac{{10}}{3}\times \displaystyle \frac{7}{6}\\ \ \ \ \ x=14\\\\ \text{(b)}\;\;{{\log }_{b}}x=2-a+{{\log }_{b}}\left( {\displaystyle \frac{{{{a}^{2}}{{b}^{a}}}}{{{{b}^{2}}}}} \right)\\ \ \ \ \ {{\log }_{b}}x=2-a+{{\log }_{b}}{{a}^{2}}+{{\log }_{b}}{{b}^{a}}-{{\log }_{b}}{{b}^{2}}\\ \ \ \ \ {{\log }_{b}}x=2-a+{{\log }_{b}}{{a}^{2}}+a-2\\ \ \ \ \ {{\log }_{b}}x={{\log }_{b}}{{a}^{2}}\\ \ \ \ \ x={{a}^{2}}\\\\ \text{(c)}\;\;\log {{x}^{3}}-\log {{x}^{2}}=\log 5x-\log 4x\\ \ \ \ \ \log \displaystyle \frac{{{{x}^{3}}}}{{{{x}^{2}}}}=\log \displaystyle \frac{{5x}}{{4x}}\\ \ \ \ \ \log x=\log \displaystyle \frac{5}{4}\\ \ \ \ \ x=\displaystyle \frac{5}{4}\\\\ \text{(d)}\;\;{{\log }_{{10}}}x+{{\log }_{{10}}}3={{\log }_{{10}}}6\\ \ \ \ \ \ {{\log }_{{10}}}3x={{\log }_{{10}}}6\\ \ \ \ \ \ 3x=6\\ \ \ \ \ \ x=2\\ \text{(e)}\;\;8\log x=\log {{a}^{{\frac{3}{2}}}}+\log 2- \frac{1}{2}\log {{a}^{3}}-\log \displaystyle \frac{2}{{{{a}^{4}}}}\\ \ \ \ \ \log {{x}^{8}}=\log {{a}^{{\frac{3}{2}}}}+\log 2-\log {{a}^{{ \frac{3}{2}}}}-\left( {\log 2-\log {{a}^{4}}} \right)\\ \ \ \ \ \log {{x}^{8}}=\log {{a}^{4}}\\ \ \ \ \ {{x}^{8}}={{a}^{4}}\\ \ \ \ \ x=\sqrt{a}\end{array}$

7.           Given that $\log_{10} 5 = 0.6990$ and $\log_{10}x = 0.2330$. What is the value of $x$?

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$\begin{array}{l} \ \ \;\;{{\log }_{{10}}}5=0.6990\\\\ \ \ \ \ {{\log }_{{10}}}x=0.2330\\\\ \ \ \ \ \therefore \ \ 3{{\log }_{{10}}}x=0.6990\\\\ \ \ \ \ \therefore \ \ 3{{\log }_{{10}}}x={{\log }_{{10}}}5\\\\ \ \ \ \ \therefore \ \ {{\log }_{{10}}}x=\displaystyle\frac{1}{3}{{\log }_{{10}}}5\\\\ \ \ \ \ \therefore \ \ {{\log }_{{10}}}x={{\log }_{{10}}}{{5}^{{\frac{1}{3}}}}\\\\ \ \ \ \ \therefore \ \ x={{5}^{{\frac{1}{3}}}}=\sqrt[3]{5} \end{array}$

8.           Show that if $\log _{e} I=-\displaystyle\frac{R}{L} t+\log _{e} I_{0}$ then $I=I_{0} e^{-\frac{R t}{L}}$

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$\begin{array}{l} {{\log }_{e}}I=-\displaystyle \frac{R}{L}t+{{\log }_{e}}{{I}_{0}}\\\\ {{\log }_{e}}I-{{\log }_{e}}{{I}_{0}}=-\displaystyle \frac{{Rt}}{L}\\\\ {{\log }_{e}}\displaystyle \frac{I}{{{{I}_{0}}}}=-\displaystyle \frac{{Rt}}{L}\\\\ \displaystyle \frac{I}{{{{I}_{0}}}}={{e}^{{-\frac{{Rt}}{L}}}}\\\\ \therefore \ I={{I}_{0}}{{e}^{{-\frac{{Rt}}{L}}}} \end{array}$

9.           Show that if $\log _{b} y=\displaystyle\frac{1}{2} \log _{b} x+c$ then $y=b^{c} \sqrt{x}$

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$\begin{array}{l}{{\log }_{b}}y=\displaystyle \frac{1}{2}{{\log }_{b}}x+c\\\\ {{\log }_{b}}y-\displaystyle \frac{1}{2}{{\log }_{b}}x=c\\\\ {{\log }_{b}}y-{{\log }_{b}}{{x}^{{\frac{1}{2}}}}=c\\\\ {{\log }_{b}}\displaystyle \frac{y}{{\sqrt{x}}}=c\\\\ \displaystyle \frac{y}{{\sqrt{x}}}={{b}^{c}}\\\\y=\ {{b}^{c}}\sqrt{x} \end{array}$

10.           Show that

$\begin{array}{l} \text{(a)}\ \ \displaystyle\frac{1}{4} \log _{10} 8+\frac{1}{4} \log _{10} 2=\log _{10} 2\\\\ \text{(b)}\ \ 4 \log _{10} 3-2 \log _{10} 3+1=\log _{10} 90 \end{array}$

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$\begin{array}{l}\text{(a)}\;\;\ \ \ \displaystyle \frac{1}{4}{{\log }_{{10}}}8+\displaystyle \frac{1}{4}{{\log }_{{10}}}2\\\ \ \ \ =\displaystyle \frac{1}{4}\left( {{{{\log }}_{{10}}}8+{{{\log }}_{{10}}}2} \right)\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}\left( {8\times 2} \right)\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}16\\\ \ \ \ =\displaystyle \frac{1}{4}{{\log }_{{10}}}{{2}^{4}}\\\ \ \ \ =\displaystyle \frac{1}{4}\times 4{{\log }_{{10}}}2\\\ \ \ \ ={{\log }_{{10}}}2\\\\\text{(b)}\;\;\ \ 4{{\log }_{{10}}}3-2{{\log }_{{10}}}3+1\\\ \ \ \ =2{{\log }_{{10}}}3+{{\log }_{{10}}}10\\\ \ \ \ ={{\log }_{{10}}}{{3}^{2}}+{{\log }_{{10}}}10\\\ \ \ \ ={{\log }_{{10}}}\left( {{{3}^{2}}\times 10} \right)\\\ \ \ \ ={{\log }_{{10}}}90\end{array}$

11.           Show that

$\begin{array}{l} \text{(a)}\ \ a^{2 \log _{a} 3}+b^{3 \log _{b} 2}=17\\\\ \text{(b)}\ \ 3 \log _{6} 1296=2 \log _{4} 4096 \end{array}$

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$\begin{array}{l}\text{(a)}\;\;\ \ {{a}^{{2{{{\log }}_{a}}3}}}+{{b}^{{3{{{\log }}_{b}}2}}}\\\ \ \ \ ={{a}^{{{{{\log }}_{a}}{{3}^{2}}}}}+{{b}^{{{{{\log }}_{b}}{{2}^{3}}}}}\\\ \ \ \ ={{3}^{2}}+{{2}^{3}}\\\ \ \ \ =17\\\\\text{(b)}\;\ \ 3{{\log }_{6}}1296=3{{\log }_{6}}{{6}^{4}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =3\times 4{{\log }_{6}}6\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =12\\\ \ \ \ \ \ 2{{\log }_{4}}4096=2{{\log }_{4}}{{4}^{6}}\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =2\times 6{{\log }_{4}}4\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =12\\\ \ \ \therefore \ \ 3{{\log }_{6}}1296=2{{\log }_{4}}4096\end{array}$

12.           Given that $\log _{10} 12=1.0792$ and $\log _{10} 24=1.3802,$ deduce the values of $\log _{10} 2$ and $\log _{10} 6$.

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$\begin{array}{l}{{\log }_{{10}}}12=1.0792\\\\{{\log }_{{10}}}24=1.3802\\\\{{\log }_{{10}}}2={{\log }_{{10}}}\displaystyle \frac{{24}}{{12}}\\\ \ \ \ \ \ \ ={{\log }_{{10}}}24-{{\log }_{{10}}}12\\\ \ \ \ \ \ \ =1.3802-1.0792\\\ \ \ \ \ \ \ =0.3010\\\\{{\log }_{{10}}}6={{\log }_{{10}}}\displaystyle \frac{{12}}{2}\\\ \ \ \ \ \ \ ={{\log }_{{10}}}12-{{\log }_{{10}}}2\\\ \ \ \ \ \ \ =1.0792-0.3010\\\ \ \ \ \ \ \ =0.7782\end{array}$

13.           If $\log _{x} a=5$ and $\log _{x} 3 a=9$, find the values of $a$ and $x$.

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$\begin{array}{l}{{\log }_{x}}a=5\\\\\therefore \ \ a={{x}^{5}}\\\\{{\log }_{x}}3a=9\\\\\therefore \ \ 3a={{x}^{9}}\\\\\therefore \ \ 3{{x}^{5}}={{x}^{9}}\\\\\therefore \ \ {{x}^{4}}=3\\\\\therefore \ \ x={{3}^{{\frac{1}{4}}}}=\sqrt[4]{3}\\\\\therefore \ \ a={{\left( {{{3}^{{\frac{1}{4}}}}} \right)}^{5}}={{3}^{{\frac{5}{4}}}}=3\sqrt[4]{3}\end{array}$

14.           $\text { (a) }$ If $\log _{10} 2=a,$ find $\log _{10} 8+\log _{10} 25$ in terms of $a$.

$\text { (b) }$ If $a=10^{x}$ and $b=10^{y},$ express $\log _{10}\left(a^{4} b^{3}\right)$ in terms of $x$ and $y$.

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$\begin{array}{l}(\text{a})\ {{\log }_{{10}}}2=a\ (\text{given})\\\\\ \ \ {{\log }_{{10}}}8+{{\log }_{{10}}}25\\\\={{\log }_{{10}}}8+{{\log }_{{10}}}\frac{{100}}{4}\\\\={{\log }_{{10}}}8+{{\log }_{{10}}}100-{{\log }_{{10}}}4\\\\={{\log }_{{10}}}{{2}^{3}}+{{\log }_{{10}}}{{10}^{2}}-{{\log }_{{10}}}{{2}^{2}}\\\\=3{{\log }_{{10}}}2+2{{\log }_{{10}}}10-2{{\log }_{{10}}}2\\\\={{\log }_{{10}}}2+2\\\\=a+2\\\\(\text{b})\ \ \left. \begin{array}{l}a={{10}^{x}}\\b={{10}^{y}}\end{array} \right\}(\text{given})\\\\\ \ \ \ \ {{\log }_{{10}}}\left( {{{a}^{4}}{{b}^{3}}} \right)={{\log }_{{10}}}\left( {{{{\left( {{{{10}}^{x}}} \right)}}^{4}}{{{\left( {{{{10}}^{y}}} \right)}}^{3}}} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}\left( {\left( {{{{10}}^{{4x}}}} \right)\left( {{{{10}}^{3}}^{y}} \right)} \right)\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ={{\log }_{{10}}}{{10}^{{4x+3y}}}\\\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =4x+3y\end{array}$

15.           $\text { (a) }$ If $\log _{2}(4 x-4)=2$, find the value of $\log _{4} x$.

$\text { (b) }$ Prove that if $\displaystyle\frac{1}{2} \log _{3} M+3 \log _{3} N=1$ then $M N^{6}=9$.

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$\begin{array}{l} (\text{a})\ {{\log }_{2}}(4x-4)=2\\\\ \ \ \ 4x-4={{2}^{2}}\\\\ \ \ \ 4x=8\\\\ \ \ \ x=2\\\\ \ \ \ x={{4}^{{\frac{1}{2}}}}\\\\ \ \ \ {{\log }_{4}}x=\displaystyle\frac{1}{2}\\\\\\ (\text{b})\ \ \displaystyle\frac{1}{2}{{\log }_{3}}M+3{{\log }_{3}}N=1\\\\ \ \ \ \ \ {{\log }_{3}}{{M}^{{\frac{1}{2}}}}+{{\log }_{3}}{{N}^{3}}=1\\\\ \ \ \ \ \ {{\log }_{3}}\left( {{{M}^{{\frac{1}{2}}}}{{N}^{3}}} \right)=1\\\\ \ \ \ \ \ {{M}^{{\frac{1}{2}}}}{{N}^{3}}=3\\\\ \ \ \ \ \ \text{Squaring both sides}\text{.}\\\\ \ \ \ \ \ M{{N}^{6}}=9 \end{array}$

• တိုင်းတာမှုများပါသော တွက်ချက်မှုများ ဆောင်ရွက်ရာတွင် ရလဒ်ကို တိကျမှုအနည်းဆုံး ဖြစ်သော အတိုင်းအတာပမာဏအတိုင်း ဖေါ်ပြပေးသင့်ပါသည်။

• ပေါင်းခြင်း နှင့် နုတ်ခြင်းတွင် ရလဒ်ကို တိကျမှုအနည်းဆုံး ဖြစ်သော အတိုင်းအတာ၏ ဒသမ အရေအတွက်အတိုင်း ဖေါ်ပြပေးသင့်ပါသည်။

• မြှောက်ခြင်းနှင့် စားခြင်းတွင် ရလဒ်ကို တိကျမှုအနည်းဆုံး ဖြစ်သော အတိုင်းအတာ၏ အရာရောက်ဂဏန်း အရေအတွက်အတိုင်း ဖေါ်ပြပေးသင့်ပါသည်။

• #### The Rules for Significant Figures

$\begin{array}{|l|l|l|l|} \hline \text { Example } & \begin{array}{c} \text { Scientific } \\ \text { Notation } \end{array} & \begin{array}{c} \text { Number of } \\ \text { Significant } \\ \text { Figures } \end{array} & {\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text { Remark }} \\ \hline 0.00682 & 6.82 \times 10^{-3} & \ \ \ \ \ \ \ \ 3 & \text { Leading zeros are not significant. } \\ \hline 100.07 & 1.0007 \times 10^{2} & \ \ \ \ \ \ \ \ 5 & \text { Embedded zeros are always significant. } \\ \hline 300 & 3 \times 10^{2} & \ \ \ \ \ \ \ \ 1 & \begin{array}{l} \text { Trailing zeros are significant only if the } \\ \text { decimal point is specified. If the decimal } \\ \text { point is not specified, trailing zeros are } \\ \text { generally not significant. } \end{array} \\ \hline 300 . & 3.00 \times 10^{2} & \ \ \ \ \ \ \ \ 3 & \text { All digits are significant. } \\ \hline 300.0 & 3.000 \times 10^{3} & \ \ \ \ \ \ \ \ 4 & \text { All digits are significant. } \\ \hline 0.0080500 & \begin{array}{l} 8.0500 \times10^{-3} \\ \end{array} & \ \ \ \ \ \ \ \ 5 & \begin{array}{l} \text { Leading zeros are not significant whereas } \\ \text { embedded zeros and trailing zeros after } \\ \text { the decimal point are significant. } \end{array} \\ \hline \end{array}$

1.           How many significant figures are there in each of the following numbers?

$\begin{array}{lll} \text{(a)}\ \ 2.175 & \text{(b)}\ \ 0.2175& \text{(c)}\ \ 0.0075\\\\ \text{(d)}\ \ 89400 & \text{(e)}\ \ 0.00046 & \text{(f)}\ \ 12.0500 & \end{array}$

Show/Hide Solution

$\begin{array}{lll} \text{(a)}\ \ 2.175 & \Rightarrow & 4\ \text{significant figures}\\\\ \text{(b)}\ \ 0.2175 & \Rightarrow & 4\ \text{significant figures}\\\\ \text{(c)}\ \ 0.0075 & \Rightarrow & 2\ \text{significant figures}\\\\ \text{(d)}\ \ 89400 & \Rightarrow & 3\ \text{significant figures}\\\\ \text{(e)}\ \ 0.00046 & \Rightarrow & 2\ \text{significant figures}\\\\ \text{(f)}\ \ 12.0500 & \Rightarrow & 6\ \text{significant figures}\\\\ \end{array}$

2.           Write in scientific notation.

$\begin{array}{lll} \text{(a)}\ \ 24.86 & \text{(b)}\ \ 2.486& \text{(c)}\ \ 0.2486\\\\ \text{(d)}\ \ 0.002486 & \text{(e)}\ \ 0.073 & \text{(f)}\ \ 0.0086\\\\ \text{(g)}\ \ 0.934 & \text{(h)}\ \ 7 & \text{(i)}\ \ 6.843250\\\\ \text{(j)}\ \ 0.00056857 & \text{(k)}\ \ 62500 & \text{(l)}\ \ 3001 \end{array}$

Show/Hide Solution

$\begin{array}{lll} \text{(a)}\ \ 24.86 & = & 2.486\times 10^1 \\\\ \text{(b)}\ \ 2.486 & = & 2.486\times 10^0 \\\\ \text{(c)}\ \ 0.2486 & = & 2.486\times 10^{-1} \\\\ \text{(d)}\ \ 0.002486 & = & 2.486\times 10^{-3} \\\\ \text{(e)}\ \ 0.073 & = & 7.3\times 10^{-2} \\\\ \text{(f)}\ \ 0.0086 & = & 8.6\times 10^{-3} \\\\ \text{(g)}\ \ 0.934 & = & 9.34\times 10^{-1} \\\\ \text{(h)}\ \ 7 & = & 7\times 10^0 \\\\ \text{(i)}\ \ 6.843250 & = & 6.843250\times 10^0 \\\\ \text{(j)}\ \ 0.00056857& = & 5.6857\times 10^{-4} \\\\ \text{(k)}\ \ 62500 & = & 6.25\times 10^{4} \\\\ \text{(l)}\ \ 3001 & = & 3.001\times 10^3 \end{array}$

3.           Write each number in ordinary decimal form.

$\begin{array}{l} \text{(a)}\ \ 7.84 \times 10^{4}\\\\ \text{(b)}\ \ 7.89 \times 10^{-4}\\\\ \text{(c)}\ \ 2.25 \times 10^{5}\\\\ \text{(d)}\ \ 4.01 \times 10^{-3} \end{array}$

Show/Hide Solution

$\begin{array}{lll} \text{(a)}\ \ 7.84 \times 10^{4} &=& 78400\\\\ \text{(b)}\ \ 7.89 \times 10^{-4} &=& 0.000789\\\\ \text{(c)}\ \ 2.25 \times 10^{5} &=& 225000\\\\ \text{(d)}\ \ 4.01 \times 10^{-3} &=& 0.00401 \end{array}$

4.           Simplify and give the answers in scientific notation.

$\begin{array}{l} \text{(a)}\ \ 2.3 \times 10^{2}+1.7 \times 10^{2}\\\\ \text{(b)}\ \ 4.6 \times 10^{-3}-2.5 \times 10^{-3}\\\\ \text{(c)}\ \ \left(4.5 \times 10^{6}\right) \times\left(1.5 \times 10^{-2}\right)\\\\ \text{(d)}\ \ \displaystyle \frac{7.6 \times 10^{5}}{1.9 \times 10^{-2}} \end{array}$

Show/Hide Solution

$\begin{array}{l} \text{(a)}\;\;\ 2.3\times {{10}^{2}}+1.7\times {{10}^{2}}\\ \ \ \ =(2.3+1.7)\times {{10}^{2}}\\ \ \ \ =4.0\times {{10}^{2}}\\\\ \text{(b)}\;\;\ 4.6\times {{10}^{{-3}}}-2.5\times {{10}^{{-3}}}\\ \ \ \ =(4.6-2.5)\times {{10}^{{-3}}}\\ \ \ \ =2.1\times {{10}^{{-3}}}\\\\ \text{(c)}\;\;\;\left( {4.5\times {{{10}}^{6}}} \right)\times \left( {1.5\times {{{10}}^{{-2}}}} \right)\\ \ \ \ =(4.5\times 1.5)\times {{10}^{{6-2}}}\\ \ \ \ =6.8\times {{10}^{4}}\\\\ \text{(d)}\;\;\;\displaystyle \frac{{7.6\times {{{10}}^{5}}}}{{1.9\times {{{10}}^{{-2}}}}}\\ \ \ \ =4.0\times {{10}^{{5+2}}}\\ \ \ \ =4.0\times {{10}^{7}} \end{array}$

5.           Compute using scientific notation.

$\begin{array}{l} \text{(a)}\ \ \displaystyle \frac{2.5 \times 10^{2}}{0.25 \times 0.002}\\\\ \text{(b)}\ \ \displaystyle \frac{33,000,000 \times 0.4}{1.1 \times 30}\\\\ \text{(c)}\ \ \displaystyle \frac{50 \times 0.014 \times 0.30}{10500}\\\\ \text{(d)}\ \ \displaystyle \frac{7000 \times 80 \times 300}{400} \end{array}$

Show/Hide Solution

$\begin{array}{l}\text{(a)}\;\;\ \displaystyle \frac{{2.5\times {{{10}}^{2}}}}{{0.25\times 0.002}}\\\ \ \ =\displaystyle \frac{{2.5\times {{{10}}^{2}}}}{{2.5\times {{{10}}^{{-1}}}\times 2.0\times {{{10}}^{{-3}}}}}\\\ \ \ =\displaystyle \frac{1}{2}\times {{10}^{{2+1+3}}}\\\ \ \ =0.5\times {{10}^{6}}\\\ \ \ =5.0\times {{10}^{5}}\\\\\text{(b)}\;\;\ \displaystyle \frac{{33,000,000\times 0.4}}{{1.1\times 30}}\\\ \ \ =\displaystyle \frac{{3.3\times {{{10}}^{7}}\times 4.0\times {{{10}}^{{-1}}}}}{{1.1\times 3.0\times {{{10}}^{1}}}}\\\ \ \ =\displaystyle \frac{{3.3\times 4.0}}{{3.3}}\times {{10}^{{7-1-1}}}\\\ \ \ =4.0\times {{10}^{5}}\\\\\text{(c)}\;\;\;\displaystyle \frac{{50\times 0.014\times 0.30}}{{10500}}\\\ \ \ =\displaystyle \frac{{5.0\times {{{10}}^{1}}\times 1.4\times {{{10}}^{{-2}}}\times 3.0\times {{{10}}^{{-1}}}}}{{1.05\times {{{10}}^{4}}}}\\\ \ \ =\displaystyle \frac{{5.0\times 1.4\times 3.0}}{{1.05}}\times {{10}^{{1-2-1-4}}}\\\ \ \ =\displaystyle \frac{{2.1\times 10}}{{1.05}}\times {{10}^{{-6}}}\\\ \ \ =2\times {{10}^{{-5}}}\\\\\text{(d)}\;\;\;\displaystyle \frac{{7000\times 80\times 300}}{{400}}\\\ \ \ =\displaystyle \frac{{7\times {{{10}}^{3}}\times 8\times {{{10}}^{1}}\times 3\times {{{10}}^{2}}}}{{4\times {{{10}}^{2}}}}\\\ \ \ =\displaystyle \frac{{7\times 8\times 3}}{4}\times {{10}^{{3+2+1-2}}}\\\ \ \ =42\times {{10}^{4}}\\\ \ \ =4.2\times {{10}^{5}}\end{array}$