# 39 QUESTIONS ON LOGARITHM AND LOGARITHMIC EQUATIONS

Logarithms are the inverse function of exponentiation.

Different rules guide the operation of logarithms, let us take a closer look at them

1. $\log_xa+\log_xb=\log_xab$

2. $\log_xa-\log_xb=\log_x\frac{a}{b}$

3. $\log_xa^b=b\log_xa$

4. $\log_xa=\frac{\log_ba}{\log_bx}$

5. $\log_x1=0$

We would be applying these rules as we solved these 39 questions and answers on the logarithm.

The purpose of these 30 logarithm questions and answers is to aid students in their comprehension of logarithms and logarithmic equations.

Forthwith, Let's get started

#### Question 1

Solve $\log_{10}6+\log_{10}{45}-\log_{10}27$

Solution:

Recalled that $\log_ya+\log_y b=\log_y ab$

$\log_{10}\frac{6 ×45}{27}$

$\log_{10}10=1$

#### Question 2

Solve for q if $\log_{10}q=2.7078$

Solution:

Converting to index form

$q=10^{2.7078}$

$q=510.2$

#### Question 3

Solve for a in the equation $3\log a+ 5\log a-6\log a=\log 64$

Solution:

Recalled that $b\log a=\log a^b$

$\log a^3+ \log a^5-\log a^6=\log 64$

$\log \frac{a^3 × a^5}{a^6}=\log 64$

By the law of indices

$\log a^{3+5-6}=\log 64$

$\log a^{2}=\log 64$

Since the log base is the same,

$a^{2}=64$

$\sqrt{a^2}=\sqrt{64}$

$a=8$

#### Question 4

$6\log (x+4)=\log 64$

Solution:

$\log (x+4)^6=\log 64$

$(x+4)^6=2^6$

$\sqrt[6]{(x+4)^6}=\sqrt[6]{2^6}$

$x+4=2$

$x=2-4=-2$

#### Question 5

Solve for p in the equation $\frac{1}{3}\log_{10}p=1$

Solution:

$\log_{10}p^{ \frac{1}{3}}=1$

Converting to index form

$p^{\frac{1}{3}}=10^1$

$\sqrt[3]{p}=10$

Add a cube to both sides

$\left(\sqrt[3]{p}\right)^3=10^3$

$p=1000$

#### Question 6

Simplify $\frac{\log \sqrt{8}}{\log 8}$

Solution:

$\frac{\log 8^{\frac{1}{2}}}{\log 8}$

$\frac{ \frac{1}{2}\require{cancel}\bcancel{\log 8}}{\require{cancel}\bcancel{\log 8}}=\frac{1}{2}$

#### Question 7

Evaluate $\log 6+\log 2 -\log 12$

Solution:

$\log \frac{6 \times 2}{12}$

$\log 1$

$\log_{10}1=0$

Note: Whenever we are not given a base, we automatically assume it is base 10.

#### Question 8

Evaluate $\frac{\log 81}{\log \frac{1}{3}}$

Solution:

The above can be rewritten as

$\frac{\log 3^4}{\log 3^{-1}}$

$\frac{4\require{cancel}\bcancel{\log 3}}{-1\require{cancel}\bcancel{\log 3}}=-4$

#### Question 9

If $\log_{10}(2x+1)-\log_{10}(3x-2)=1$

Solution:

$\log_{10}\frac{2x+1}{3x-2}=1$

Converting to index form.

$\frac{2x+1}{3x-2}=10$

$2x+1=30x-20$

$-28x=-21$

$x=\frac{3}{4}$

#### Question 10

Simplify $\frac{\log \sqrt{27}}{\log \sqrt{81}}$

Solution:

$\frac{\log 27^{\frac{1}{2}}}{81^{\frac{1}{2}}}$

$\frac{\log 3^{3\left(\frac{1}{2}\right)}}{3^{4\left(\frac{1}{2}\right)}}$

$\frac{\log 3^{\frac{3}{2}}}{3^{2}}$

$\frac{\frac{3}{2}\log 3}{2\log 3}$

$\frac{\frac{3}{2}}{2}=\frac{3}{4}$

#### Question 11

$\log_{10}25+\log_{10}32-\log_{10}8$

Solution:

$\log_{10}\frac{25 × 32}{8}$

$\log_{10}\frac{800}{8}$

$\log_{10}100=2$

#### Question 12

Find x if $\log_{10}(2x+1)-\log_{10}(3x-2)=1$

Solution:

$\log_{10}\frac{2x+1}{3x-2}=1$

Converting to index form

$\frac{2x+1}{3x-2}=10$

$2x+1=10(3x-2)$

$2x+1=30x-20$

$2x-30x=-20-1$

$-28x=-21$

$x=\frac{3}{4}$

#### Question 13

$\log_{10}\frac{30}{16}-2\log_{10}\frac{5}{9}+\log_{10}\frac{400}{243}$

Solution:

$\log_{10}\frac{30}{16}-\log_{10}\left(\frac{5}{9}\right)^2+\log_{10}\frac{400}{243}$

$\log_{10}\frac{30}{16}-\log_{10}\frac{25}{81}+\log_{10}\frac{400}{243}$

$\log_{10}\frac{\frac{30}{16}}{\frac{25}{81}}\times\frac{400}{243}$

$\log_{10}\frac{30}{16}\times\frac{81}{25}\times\frac{400}{243}$

$\log_{10}{10}=1$

#### Question 14

Evaluate $\log_{10}\sqrt{35}+\log_{10}\sqrt{2}-\log_{10}\sqrt{7}$

Solution:

$\log_{10}\frac{\sqrt{35} ×\sqrt{2}}{\sqrt{7}}$

$\log_{10}\frac{\sqrt{70}}{\sqrt{7}}$

$\log_{10}\frac{\sqrt{10} ×\require{cancel}\bcancel{\sqrt{7}}}{\require{cancel}\bcancel{\sqrt{7}}}$

$\log_{10}10^{\frac{1}{2}}$

$\frac{1}{2}\log_{10}{10}=\frac{1}{2}$

#### Question 15

If $\log_{10}(3x-1)-\log_{10}2=3$

Solution:

$\log_{10}\frac{3x-1}{2}=3$

Converting to index form

$\frac{3x-1}{2}=10^{3}$

$\frac{3x-1}{2}=1000$

$3x-1=2000$

$3x=2001$

$x=667$

#### Question 16

Find a given that $\log_2a=\log_84$

Solution:

Recalled that $\log_ba=\frac{\log_ca}{\log_cb}$

$\log_2a=\frac{\log_24}{\log_28}$

$\log_2a=\frac{2\log_22}{3\log_22}$

$\log_2a=\frac{2}{3}$

Converting to index form

$a=2^{\frac{2}{3}}$

#### Question 17

$\frac{1}{2}\log_{10}{p}=1$

Solution:

$\log_{10}p^{\frac{1}{2}}=1$

$p^{\frac{1}{2}}=10$

$\sqrt{p}=10$

Add a square to both sides

$\left(\sqrt{p}\right)^2=(10)^2$

$p=100$

#### Question 18

Solve the simultaneous equation

$\log_{10}x+\log_{10}y=4$, $\log_{10}x+2\log_{10}y=3$

Solution:

$\log_{10}x+\log_{10}y=4$

$\log_{10}x+\log_{10}y^2=3$

$\log_{10}xy=4$

$\log_{10}xy^2=3$

Recalled that $\log_{10}10=1$

$\log_{10}xy=4(\log_{10}10)$

$\log_{10}xy^2=3(\log_{10}10)$

$\log_{10}xy=\log_{10}10^4$

$\log_{10}xy^2=\log_{10}10^3$

$\log_{10}xy=\log_{10}10000$

$\log_{10}xy^2=\log_{10}1000$

Since both sides have same base

$xy=10000$.....eqn 1

$xy^2=1000$.....eqn 2

Divide eqn 1 by eqn 2

$\begin{matrix} xy=10000\\ ÷(xy^2=1000)\\ \hline y^{1-2}=10 \end{matrix}$

$y^{-1}=10$

$\frac{1}{y}=10$

$1=10y$

$y=\frac{1}{10}$

Inserting the value of y in eqn 1

$x(\frac{1}{10})=10000$

$\frac{x}{10}=10000$

$x=100000$

#### Question 19

Evaluate $\log_{10}35$ given that $\log_{10}7=0.8451$ and $\log_{10}5=0.6990$

Solution:

$\log_{10}35=\log_{10}7×5=\log_{10}7+\log_{10}5$

Remember that $\log_{10}7=0.8451$ and $\log_{10}5=0.6990$

$\log_{10}35=0.8451+0.6990=1.5441$

#### Question 20

Evaluate $\log_5\frac{3}{5}+3\log_5\frac{15}{2}-\log_5\frac{81}{8}$

Solution:

$\log_5\frac{3}{5}+\log_5\left(\frac{15}{2}\right)^3-\log_5\frac{81}{8}$

$\log_5\frac{3}{5}+\log_5\frac{3375}{8}-\log_5\frac{81}{8}$

$\log_5\frac{\frac{3}{5} \times \frac{3375}{8}}{\frac{81}{8}}$

$\log_5\frac{\frac{10125}{40}}{\frac{81}{8}}$

$\log_5\frac{10125}{40} ×\frac{8}{81}$

$\log_525=2$

#### Question 21

Evaluate $\log_{10}0.24$ given that $\log_{10} 2=0.3010$ and $\log_{10}3=0.4771$

Solution:

$\log_{10}\frac{24}{100}$

$\log_{10}24-\log_{10}{100}$

$\log_{10}8 ×3-2(\log_{10}{10})$

$\log_{10}8+\log_{10}3-2$

$\log_{10}2^3+\log_{10}3-2$

Recalled that $\log_{10} 2=0.3010$ and $\log_{10}3=0.4771$

$3(0.3010)+0.4771-2=-0.6199$

#### Question 22

Evaluate $\frac{\log_28}{\log_2\frac{1}{4}}$

Solution:

$\frac{\log_22^3}{\log_22^{-2}}$

$\frac{3\log_22}{-2\log_22}$

$\frac{3(1)}{-2(1)}=\frac{3}{-2}$

#### Question 23

Evaluate $\frac{\log_2\sqrt{8}}{\log_24-\log_22}$

Solution:

$\frac{\log_28^{\frac{1}{2}}}{\log_2\frac{4}{2}}$

$\frac{\log_22^{\frac{3}{2}}}{\log_22}$

$\frac{\frac{3}{2}(1)}{1}=\frac{3}{2}$

#### Question 24

Evaluated $\log \left(\frac{35 \times 49}{40÷56}\right)$ given that $\log 5=0.6990$, $\log 7=0.8451$

Solution:

$\left(\log 35 × 49\right)-\left(\log 40 ÷ 56\right)$

$\log 35+\log 49-\left(\log \frac{40}{56}\right)$

$\log 7×5 +\log 7^2-\left(\log \frac{5} {7}\right)$

$\log 7+\log 5 +2\log 7-\left(\log 5-\log 7\right)$

$0.8451+0.6990+2(0.8451)-\left(0.6990-0.8451\right)=3.3804$

#### Question 25

Evaluate $x\log_28$ given that $\log_x64=3$

Solution:

First, we need to fire in the value of x

$\log_x64=3$

Index form

$64=x^3$

$4^3=x^3$

$x=4$

$4\log_28$

$4\log_22^3$

$4×3\log_22$

$12(1)=12$

#### Question 26

Evaluate $4\log_2\frac{1}{4}-3\log_{\frac{1}{2}}32$

Solution:

$4\log_22^{-2}-3\log_{\frac{1}{2}}\left(\frac{1}{2}\right)^{-5}$

$(4×-2)\log_22-(3)(-5)\log_{\frac{1}{2}}\left(\frac{1}{2}\right)$

$-8(1)+15(1)=7$

#### Question 27

Solve for x in $\log_54x+7-\log_5x=2$

Solution:

$\log_5\frac{4x+7}{x}=2$

$\frac{4x+7}{x}=25$

$4x+7=25x$

$25x-4x=7$

$x=\frac{1}{3}$

#### Question 28

Solve for x in $\log_2x^2+4x+3=4+\log_2x^2+x$

Solution:

$\log_2x^2+4x+3-\log_2x^2+x=4$

$\log_2\frac{x^2+4x+3}{x^2+x}=4$

Converting to index form

$\frac{x^2+4x+3}{x^2+x}=2^4$

$\frac{(x+1)(x+3)}{x(x+1)}=16$

$\frac{x+3}{x}=16$

$x+3=16x$

$15x=3$

$x=\frac{1}{5}$

#### Question 29

Solve for x in the equation

$\log_x{16}=\log_x9+2$

Solution:

$\log_x{16}-\log_x9=2$

$\log_x{16}-\log_x9=2$

$\log_x\frac{16}{9}=2$

$\frac{16}{9}=x^2$

$\sqrt{\frac{16}{9}}=\sqrt{x^2}$

$x=\frac{4}{3}$

#### Question 30

$5 ×5^{\log x}+5^{2-\log x}=30$

Solution:

Using law of index, the above can be re-written  as:

$5 x 5^{\log x}+\frac{5^2}{5^{\log x}}=30$

Let $5^{\log x}=b$

$5×b+\frac{25}{b}=30$

$\frac{5b^2+25}{b}=30$

$5b^2+25=30b$

$5b^2-30b+25=0$

If we divide through by 5, the above can be rewritten as

$b^2-6b-5=0$

$b(b-1)-5(b-1)=0$

$b=5$ or $b=1$

Recall that $b=5^{\log x}$

when b=5

$5^{\log x}=5^1$

$\log x=1$

Convert to index form

$x=10^1=10$

when b=1

$5^{\log x}=5^0$

$\log x=0$

Convert to index form

$x=10^0=1$

#### Question 31

Simplify $2\log_{10}(x+a)-\log_{10}(x^2-a^2)$

Solution:

$\log_{10}(x+a)^2-\log_{10}(x^2-a^2)$

$\log_{10}\frac{(x+a)^2}{x^2-a^2}$

Recall difference of two square,

$$a^2-b^2=(a+b)(a-b)$$

$\log_{10}\frac{(x+a)(x+a)}{(x+a)(x-a)}$

$\log_{10}\frac{x+a}{x-a}$

#### Question 32

Evaluate $5^\left({\log_5x^3}\right)$

Solution:

Let the result of $5^\left({\log_5x^3}\right)=y$

Adding $\log_5$ to both sides

$\log_55^\left({\log_5x^3}\right)=\log_5y$

$\log_5x^3\left(\log_55\right)=\log_5y$

$\log_5x^3=\log_5y$

The log base is the same. So, we equate the power

$y=x^3$

Since y is the result of $5^\left({\log_5x^3}\right)$, it follow that:

$5^\left({\log_5x^3}\right)=x^3$

#### Question 33

Evaluate $\log_{10}25-\log_{10}8+\log_{10}16$

Solution:

Using calculator

$1.3979-0.9031+1.2041=1.6989$

#### Question 34

Find the value of $\log_2\left(\log_2\left(\log_3\left(\log_327^3\right)\right)\right)$

Solution:

First, we simplify from the up

$\log_2\left(\log_2\left(\log_3\left(\log_33^{3(3)}\right)\right)\right)$

$\log_2\left(\log_2\left(\log_3\left(9\log_33\right)\right)\right)$

$\log_2\left(\log_2\left(\log_39\right)\right)$

$\log_2\left(\log_22\right)$

$\log_21=0$

#### Question 35

$\log_2x+\log_4x-\log_{16}x=\frac{21}{4}$

Solution:

Remember that

$\log_ab=\frac{\log_bc}{\log_ba}$

$\log_2x+\frac{\log_2x}{\log_24}+\frac{\log_2x}{\log_216}=\frac{21}{4}$

$\log_2x+\frac{\log_2x}{2}+\frac{\log_2x}{4}=\frac{21}{4}$

Taking L.C.M

$\frac{4\log_2x+2\log_2x+\log_2x}{4}=\frac{21}{4}$

If we multiply both sides by 4 so that the denominator cancel, the result is

$4\log_2x+2\log_2x+\log_2x=21$

$7\log_2x=21$

$\log_2x=3$

Converting to index form

$x=2^3=8$

#### Question 36

$\log_24^x=\log_22^{x-3}$

Solution:

If log base are the same, equate the log

$4^x=2^{x-3}$

$2^{2x}=2^{x-3}$

$2x=x-3$

$x=-3$

#### Question 37

$\log_3(4x+1)-\log_3(3x-5)=2$

Solution:

$\log_3\frac{4x+1}{3x-5}=2$

Converting to index form

$\frac{4x+1}{3x-5}=3^2$

$\frac{4x+1}{3x-5}=9$

$4x+1=9(3x-5)$

$4x+1=27x-45$

$1+45=27x-4x$

$46=23x$

$x=2$

#### Question 38

Simplify $\log_{10}^{\frac{10^x}{x}}$

Solution:

The above can be rewritten as

$\log_{10}10^x-\log_{10}^x$

$x\log_{10}10-\log_{10}^x$

$x(1)-\log_{10}^x$

$x-\log_{10}^x$

#### Question 39

Solve $3\log_69+\log_612+\log_664-\log672$

Solution:

$\log_69^3+\log_612+\log_664-log_672$

$\log_6729+\log_612+\log_664-log_672$

$\log_6\frac{729×12×64}{72}$

$\log_67776$

$\log_66^5$

$5\log_66=5(1)=5$

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Solve for x if $\log_{10}(6x-4)-\log_{10}2=1$

2
3
4
5
see solution

$\log_{10}\frac{6x-4}{2}=1$

$\frac{6x-4}{2}=10$

$6x-4=20$

$6x=24$

x=4

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