# CAT Logarithms Questions PDF [Most Expected with solutions]

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CAT Questions on Logarithms are frequently asked in the CAT Exam. You can expect around 1-2 questions in the latest 22-question format of the CAT Quant Section. Logarithm CAT concepts are often ignored by many candidates. Questions on Logs are not very tough, and hence should not be avoided. In this article, we will look into some important Logarithm questions for CAT. These are a good source for practice; If you want to practice these questions, you can also download Logarithm questions for CAT PDF, which is completely Free.

• Logarithm for CAT – Tip 1: Questions based on Logs appear in the CAT test almost every year. It is one of the easiest topics and hence should not be avoided. Based on our analysis of the CAT Previous Papers, 1-2 questions are frequently asked from Logarithms.
• Logarithm for CAT – Tip 2: Understand the CAT Syllabus and check out the Previous Year Papers of CAT. Practising Previous year’s CAT questions on logarithms helps you understand the type of questions that appear from the Logs concepts. You can check out CAT Previous Year logarithm questions with detailed solutions here.
• Logarithm for CAT – Tip 3: The Logarithms topic does not include too many concepts to master, and hence can be learned easily. Getting yourself acquainted with the basics of Logs will help you solve the problems. You can learn all the Important Logarithm Formulas for CAT, here.
• Logarithm for CAT – Tip 4: Given the 2-hour format of the CAT, Speed becomes very important. You need to solve more questions in lesser time. Hence, choosing the easy questions becomes very crucial. Logarithms are one of those easy questions, and thus, cannot be missed on the exam. Do checkout the CAT Formula Handbook which includes all the key CAT Quant Formulas.

Question 1: If $log_3 2, log_3 (2^x – 5), log_3 (2^x – 7/2)$ are in arithmetic progression, then the value of x is equal to

a) 5

b) 4

c) 2

d) 3

Question 2: Let $u = ({\log_2 x})^2 – 6 {\log_2 x} + 12$ where x is a real number. Then the equation $x^u = 256$, has

a) no solution for x

b) exactly one solution for x

c) exactly two distinct solutions for x

d) exactly three distinct solutions for x

Question 3: If x = -0.5, then which of the following has the smallest value?

a) $2^{1/x}$

b) $1/x$

c) $1/x^2$

d) $2^x$

e) $1/\sqrt{-x}$

Question 4: If $log_y x = (a*log_z y) = (b*log_x z) = ab$, then which of the following pairs of values for (a, b) is not possible?

a) (-2, 1/2)

b) (1,1)

c) (0.4, 2.5)

d) ($\pi$, 1/ $\pi$)

e) (2,2)

Question 5: If x >= y and y > 1, then the value of the expression $log_x (x/y) + log_y (y/x)$ can never be

a) -1

b) -0.5

c) 0

d) 1

Question 6: If $f(x) = \log \frac{(1+x)}{(1-x)}$, then f(x) + f(y) is

a) $f(x+y)$

b) $f{\frac{(x+y)}{(1+xy)}}$

c) $(x+y)f{\frac{1}{(1+xy)}}$

d) $\frac{f(x)+f(y)}{(1+xy)}$

Question 7: If $\log_{2}{\log_{7}{(x^2 – x+37)}}$ = 1, then what could be the value of ‘x’?

a) 3

b) 5

c) 4

d) None of these

Question 8: Suppose, $\log_3 x = \log_{12} y = a$, where $x, y$ are positive numbers. If $G$ is the geometric mean of x and y, and $\log_6 G$ is equal to

a) $\sqrt{a}$

b) 2a

c) a/2

d) a

Question 9: The value of $\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7$ is equal to

a) 1/3

b) 2/3

c) 5/6

d) 7/6

Question 10: If x is a real number such that $\log_{3}5= \log_{5}(2 + x)$, then which of the following is true?

a) 0 < x < 3

b) 23 < x < 30

c) x > 30

d) 3 < x < 23

Question 11: If $log(2^{a}\times3^{b}\times5^{c} )$is the arithmetic mean of $log ( 2^{2}\times3^{3}\times5)$, $log(2^{6}\times3\times5^{7} )$, and $log(2 \times3^{2}\times5^{4} )$, then a equals

Question 12: If x is a positive quantity such that $2^{x}=3^{\log_{5}{2}}$. then x is equal to

a) $\log_{5}{8}$

b) $1+\log_{3}({\frac{5}{3}})$

c) $\log_{5}{9}$

d) $1+\log_{5}({\frac{3}{5}})$

Question 13: If $\log_{12}{81}=p$, then $3(\dfrac{4-p}{4+p})$ is equal to

a) $\log_{4}{16}$

b) $\log_{6}{16}$

c) $\log_{2}{8}$

d) $\log_{6}{8}$

Question 14: If $\log_{2}({5+\log_{3}{a}})=3$ and $\log_{5}({4a+12+\log_{2}{b}})=3$, then a + b is equal to

a) 59

b) 40

c) 32

d) 67

Question 15: $\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$=?

a) $\frac{1}{2}$

b) 10

c) 0

d) −4

Question 16: If p$^{3}$ = q$^{4}$ = r$^{5}$ = s$^{6}$, then the value of $log_{s}{(pqr)}$ is equal to

a) $\frac{47}{10}$

b) $\frac{24}{5}$

c) $\frac{16}{5}$

d) $1$

Question 17: Let x and y be positive real numbers such that
$\log_{5}{(x + y)} + \log_{5}{(x – y)} = 3,$ and $\log_{2}{y} – \log_{2}{x} = 1 – \log_{2}{3}$. Then $xy$ equals

a) 150

b) 25

c) 100

d) 250

Question 18: If Y is a negative number such that $2^{Y^2({\log_{3}{5})}}=5^{\log_{2}{3}}$, then Y equals to:

a) $\log_{2}(\frac{1}{5})$

b) $\log_{2}(\frac{1}{3})$

c) $-\log_{2}(\frac{1}{5})$

d) $-\log_{2}(\frac{1}{3})$

Question 19: If $\log_{4}{5}=(\log_{4}{y})(\log_{6}{\sqrt{5}})$, then y equals

Question 20: If $\log_{a}{30}=A,\log_{a}({\frac{5}{3}})=-B$ and $\log_2{a}=\frac{1}{3}$, then $\log_3{a}$ equals

a) $\frac{2}{A+B-3}$

b) $\frac{2}{A+B}-3$

c) $\frac{A+B}{2}-3$

d) $\frac{A+B-3}{2}$

Question 21: $\frac{2\times4\times8\times16}{(\log_{2}{4})^{2}(\log_{4}{8})^{3}(\log_{8}{16})^{4}}$ equals

Question 22: The value of $\log_{a}({\frac{a}{b}})+\log_{b}({\frac{b}{a}})$, for $1<a\leq b$ cannot be equal to

a) 0

b) -1

c) 1

d) -0.5

Question 23: If $5 – \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 – x} = \log_{10} \frac{1}{\sqrt{1 – x^2}}$, then 100x equals

Question 24: If $5 – \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 – x} = \log_{10} \frac{1}{\sqrt{1 – x^2}}$, then 100x equals

Question 25: If $\log_{2}[3+\log_{3} \left\{4+\log_{4}(x-1) \right\}]-2=0$ then 4x equals

$2 log (2^x – 5) = log 2 + log (2^x – 7/2)$
Let $2^x = t$
=> $(t-5)^2 = 2(t-7/2)$
=> $t^2 + 25 – 10t = 2t – 7$
=> $t^2 – 12t + 32 = 0$
=> t = 8, 4
Therefore, x = 2 or 3, but $2^x$ > 5, so x = 3

$x^u = 256$

Taking log to the base 2 on both the sides,

$u * \log_{2}{x} = \log_{2}{256}$

=>$[({\log_2 x})^2 – 6 {\log_2 x} + 12] * \log_{2}{x} = 8$

$(log_2 x)^3 – 6(log_2 x)^2 + 12log_2 x = 8$

Let $log_2 x = t$

$t^3 – 6t^2 +12t – 8 = 0$

$(t-2)^3 = 0$

Therefore, $log_2 x = 2$

=> $x = 4$ is the only solution

Hence, option B is the correct answer.

$2^p$ is always positive

$x^2$ is always non negative.

$1/\sqrt{-x}$ is always positive.

$\frac{1}{x}$ is negative when x is negative.

In this case, x is negative => $\frac{1}{x}$ is smallest.

$log_y x = ab$
$a*log_z y = ab$ => $log_z y = b$
$b*log_x z = ab$ => $log_x z = a$
$log_y x$ = $log_z y * log_x z$ => $log x/log y$ = $log y/log z * log z/log x$
=> $\frac{log x}{log y} = \frac{log y}{log x}$
=> $(log x)^2 = (log y)^2$
=> $log x = log y$ or $log x = -log y$
So, x = y or x = 1/y
So, ab = 1 or -1
Option 5) is not possible

$log_x (x/y) + log_y (y/x)$ = $1 – log_x (y) + 1 – log_y (x)$
= $2 – (log_x y + 1/log_x y)$ <= 0 (Since $log_x y + 1/log_x y$ >= 2)
So, the value of the expression cannot be 1.

If $f(x) = \log \frac{(1+x)}{(1-x)}$ then $f(y) = \log \frac{(1+y)}{(1-y)}$

Also Log (A*B)= Log A + Log B

f(x)+f(y) = $\log \frac{(1+x)(1+y)}{(1-x)(1-y)}$

=$\log\frac{\left(1+xy\ +x\ +y\right)}{\left(1+xy-x-y\right)}$

Dividing numberator and denominator by (1+xy)

$\log\frac{\frac{\left(1+xy\ +x\ +y\right)}{1+xy}}{\frac{\left(1+xy-x-y\right)}{1+xy}}$

=$\log\frac{\frac{1+xy\ }{1+xy}+\frac{\left(x+y\right)}{1+xy}}{\frac{1+xy\ }{1+xy}-\frac{\left(x+y\right)}{1+xy}}$

= $\log { \frac{1+ \frac{(x+y)}{(1+xy)}}{1- \frac{(x+y)}{(1+xy)}}}$

Hence option B.

$\log_{2}{\log_{7}{(x^2 – x+37)}}$ = 1

$\log_{7}{(x^2 – x+37)}$ = $2$

$(x^2 – x+37)$ = $7^{2}$

Given eq. can be reduced to $x^2 – x + 37 = 49$

So x can be either -3 or 4.

We know that $\log_3 x = a$ and $\log_{12} y=a$
Hence, $x = 3^a$ and $y=12^a$
Therefore, the geometric mean of $x$ and $y$ equals $\sqrt{x \times y}$
This equals $\sqrt{3^a \times 12^a} = 6^a$

Hence, $G=6^a$ Or, $\log_6 G = a$

$\log_{0.008}\sqrt{5}+\log_{\sqrt{3}}81-7$

$81 = 3^4$ and $0.008 = \frac{8}{1000} = \frac{2^{3}}{10^{3}} = \frac{1}{5^{3}} = 5^{-3}$

Hence,

$\log_{0.008}\sqrt{5}+ 8 -7$

$\log_{5^{-3}}5^{\frac{1}{2}}+ 8 -7$

$\frac{log 5^{0.5}}{log 5^{-3}} + 1$

$– \frac{1}{6} + 1$

= $\frac{5}{6}$

$1 < \log_{3}5 < 2$
=> $1 < \log_{5}(2 + x) < 2$
=> $5 < 2 + x < 25$
=> $3 < x < 23$

$log(2^{a}\times3^{b}\times5^{c} )$ = $\frac{log ( 2^{2}\times3^{3}\times5) + log(2^{6}\times3\times5^{7} ) + log(2 \times3^{2}\times5^{4} ) }{3}$

$log(2^{a}\times3^{b}\times5^{c} )$ = $\frac{log ( 2^{2+6+1}\times3^{3+1+2}\times5^{1+7+4}) }{3}$

$log(2^{a}\times3^{b}\times5^{c} )$ = $\frac{log ( 2^{9}\times3^{6}\times5^{12}) }{3}$

$3log(2^{a}\times3^{b}\times5^{c} )$ = $log ( 2^{9}\times3^{6}\times5^{12})$
Hence, 3a = 9 or a = 3

Givne that: $2^{x}=3^{\log_{5}{2}}$

$\Rightarrow$ $2^{x}=2^{\log_{5}{3}}$

$\Rightarrow$ $x=\log_{5}{3}$

$\Rightarrow$ $x=\log_{5}{\dfrac{3*5}{5}}$

$\Rightarrow$ $x=\log_{5}{5}+\log_{5}{\dfrac{3}{5}}$

$\Rightarrow$ $x=1+\log_{5}{\dfrac{3}{5}}$. Hence, option D is the correct answer.

Given that: $\log_{12}{81}=p$

$\Rightarrow$ $\log_{81}{12}=\dfrac{1}{p}$

$\Rightarrow$ $\log_{3}{3*4}=\dfrac{4}{p}$

$\Rightarrow$ $1+\log_{3}{4}=\dfrac{4}{p}$

Using Componendo and Dividendo,

$\Rightarrow$ $\dfrac{1+\log_{3}{4}-1}{1+\log_{3}{4}+1}=\dfrac{4-p}{4+p}$

$\Rightarrow$ $\dfrac{\log_{3}{4}}{2+\log_{3}{4}}=\dfrac{4-p}{4+p}$

$\Rightarrow$ $\dfrac{\log_{3}{4}}{\log_{3}{9}+\log_{3}{4}}=\dfrac{4-p}{4+p}$

$\Rightarrow$ $\dfrac{\log_{3}{4}}{\log_{3}{36}}=\dfrac{4-p}{4+p}$

$\Rightarrow$ $3*\dfrac{4-p}{4+p}=\dfrac{3\log_{3}{4}}{\log_{3}{36}}$

$\Rightarrow$ $3*\dfrac{4-p}{4+p}=\dfrac{\log_{3}{64}}{\log_{3}{36}}$

$\Rightarrow$ $3*\dfrac{4-p}{4+p}=\log_{36}{64}$

$\Rightarrow$ $3*\dfrac{4-p}{4+p}=\log_{6^2}{8^2}=\log_{6}{8}$. Hence, option D is the correct answer.

$\log_{2}({5+\log_{3}{a}})=3$
=>$5 + \log_{3}{a}$ = 8
=>$\log_{3}{a}$ = 3
or $a$ = 27

$\log_{5}({4a+12+\log_{2}{b}})=3$
=>$4a+12+\log_{2}{b}$ = 125
Putting $a$ = 27, we get
$\log_{2}{b}$ = 5
or, $b$ = 32

So, $a + b$ = 27 + 32 = 59
Hence, option A is the correct answer.

We know that $\dfrac{1}{log_{a}{b}}$ = $\dfrac{log_{x}{a}}{log_{x}{b}}$

Therefore, we can say that $\dfrac{1}{log_{2}{100}}$ = $\dfrac{log_{10}{2}}{log_{10}{100}}$

$\Rightarrow$ $\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$

$\Rightarrow$ $\dfrac{log_{10}{2}}{log_{10}{100}}$-$\dfrac{log_{10}{4}}{log_{10}{100}}$+$\dfrac{log_{10}{5}}{log_{10}{100}}$-$\dfrac{log_{10}{10}}{log_{10}{100}}$+$\dfrac{log_{10}{20}}{log_{10}{100}}$-$\dfrac{log_{10}{25}}{log_{10}{100}}$+$\dfrac{log_{10}{50}}{log_{10}{100}}$

We know that $log_{10}{100}=2$

$\Rightarrow$ $\dfrac{1}{2}*[log_{10}{2}-log_{10}{4}+log_{10}{5}-log_{10}{10}+log_{10}{20}-log_{10}{25}+log_{10}{50}]$

$\Rightarrow$ $\dfrac{1}{2}*[log_{10}{\dfrac{2*5*20*50}{4*10*25}}]$

$\Rightarrow$ $\dfrac{1}{2}*[log_{10}10]$

$\Rightarrow$ $\dfrac{1}{2}$

Given that, p$^{3}$ = q$^{4}$ = r$^{5}$ = s$^{6}$

p$^{3}$=s$^{6}$

p = s$^{\frac{6}{3}}$ = s$^{2}$   …(1)

Similarly, q = s$^{\frac{6}{4}}$ = s$^{\frac{3}{2}}$   …(2)

Similarly, r = s$^{\frac{6}{5}}$   …(3)

$\Rightarrow$ $log_{s}{(pqr)}$

By substituting value of p, q, and r from equation (1), (2) and (3)

$\Rightarrow$ $log_{s}{(s^{2}*s^{\frac{3}{2}}*s^{\frac{6}{5}})}$

$\Rightarrow$ $log_{s}(s^{\frac{47}{10}})$

$\Rightarrow$ $\dfrac{47}{10}$

Hence, option A is the correct answer.

We have, $\log_{5}{(x + y)} + \log_{5}{(x – y)} = 3$

=> $x^2-y^2=125$……(1)

$\log_{2}{y} – \log_{2}{x} = 1 – \log_{2}{3}$

=>$\ \frac{\ y}{x}$ = $\ \frac{\ 2}{3}$

=> 2x=3y   => x=$\ \frac{\ 3y}{2}$

On substituting the value of x in 1, we get

$\ \frac{\ 5x^2}{4}$=125

=>y=10, x=15

Hence xy=150

$2^{Y^2({\log_{3}{5})}}=5^{Y^2(\log_3 2)}$

Given, $5^{Y^2\left(\log_32\right)}=5^{\left(\log_23\right)}$

=> $Y^2\left(\log_32\right)=\left(\log_23\right)=>Y^2=\left(\log_23\right)^2$

=>$Y=\left(-\log_23\right)^{\ }or\ \left(\log_23\right)$

since Y is a negative number, Y=$\left(-\log_23\right)=\left(\log_2\frac{1}{3}\right)$

$\frac{\log\ 5}{2\log2}\ =\frac{\log\ y}{2\log2}\cdot\frac{\log\ 5}{2\log6}$

$\log\ 36\ =\ \log\ y;\ \therefore\ y\ =36$

$\log_a30=A\ or\ \log_a5+\log_a2+\log_a3=A$………..(1)

$\log_a\left(\frac{5}{3}\right)=-B\ or\ \log_a3-\log_a5=B$………….(2)

and finally $\log_a2=3$

Substituting this in (1) we get $\log_a5+\log_a3=A-3$

Now we have two equations in two variables (1) and (2) . On solving we get

$\log_a3=\frac{\left(A+B-3\right)}{2\ }or\ \log_3a=\frac{2}{A+B-3}$

$\frac{\left(2\cdot4\cdot8\cdot16\right)}{\left(\log_22^2\right)^2\cdot\left(\log_{2^2}2^3\right)^3\cdot\left(\log_{2^3}2^4\right)^4}\cdot$

= $\frac{2^{10}}{4\cdot\left(\frac{3}{2}\right)^3\cdot\left(\frac{4}{3}\right)^4}=24$

On expanding the expression we get $1-\log_ab+1-\log_ba$

$or\ 2-\left(\log_ab+\frac{1}{\log_ba}\right)$

Now applying the property of AM>=GM, we get that  $\frac{\left(\log_ab+\frac{1}{\log_ba}\right)}{2}\ge1\ or\ \left(\log_ab+\frac{1}{\log_ba}\right)\ge2$ Hence from here we can conclude that the expression will always be equal to 0 or less than 0. Hence any positive value is not possible. So 1 is not possible.

$5 – \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 – x} = \log_{10} \frac{1}{\sqrt{1 – x^2}}$

We can re-write the equation as: $5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\log_{10}\left(\sqrt{1+x}\times\ \sqrt{1-x}\right)^{-1}$

$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\left(-1\right)\log_{10}\left(\sqrt{1+x}\right)+\left(-1\right)\log_{10}\left(\sqrt{1-x}\right)$

$5=-\log_{10}\sqrt{1+x}+\log_{10}\sqrt{1+x}-\log_{10}\sqrt{1-x}-4\log_{10}\sqrt{1-x}$

$5=-5\log_{10}\sqrt{1-x}$

$\sqrt{1-x}=\frac{1}{10}$

Squaring both sides: $\left(\sqrt{1-x}\right)^2=\frac{1}{100}$

$\therefore\$ $x=1-\frac{1}{100}=\frac{99}{100}$

Hence, $100\ x\ =100\times\ \frac{99}{100}=99$

$5 – \log_{10}\sqrt{1 + x} + 4 \log_{10} \sqrt{1 – x} = \log_{10} \frac{1}{\sqrt{1 – x^2}}$

We can re-write the equation as: $5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\log_{10}\left(\sqrt{1+x}\times\ \sqrt{1-x}\right)^{-1}$

$5-\log_{10}\sqrt{1+x}+4\log_{10}\sqrt{1-x}=\left(-1\right)\log_{10}\left(\sqrt{1+x}\right)+\left(-1\right)\log_{10}\left(\sqrt{1-x}\right)$

$5=-\log_{10}\sqrt{1+x}+\log_{10}\sqrt{1+x}-\log_{10}\sqrt{1-x}-4\log_{10}\sqrt{1-x}$

$5=-5\log_{10}\sqrt{1-x}$

$\sqrt{1-x}=\frac{1}{10}$

Squaring both sides: $\left(\sqrt{1-x}\right)^2=\frac{1}{100}$

$\therefore\$ $x=1-\frac{1}{100}=\frac{99}{100}$

Hence, $100\ x\ =100\times\ \frac{99}{100}=99$

We have :
$\log_2\left\{3+\log_3\left\{4+\log_4\left(x-1\right)\right\}\right\}=2$
we get $3+\log_3\left\{4+\log_4\left(x-1\right)\right\}=4$
we get $\log_3\left(4+\log_4\left(x-1\right)\ =\ 1\right)$
we get $4+\log_4\left(x-1\right)\ =\ 3$
$\log_4\left(x-1\right)\ =\ -1$
x-1 = 4^-1
x = $\frac{1}{4}+1=\frac{5}{4}$
4x = 5

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