# CAT Logarithms Questions PDF [Most Important with Solutions]

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Logarithms is one of the important topics in the CAT Quants section. Over the past few years, CAT Logarithms questions have made a recurrent appearance in the Quant section. You can expect around 1-2 questions in the new format of the CAT Quant section. If you’re new to this section, you can check out the CAT Logarithms Questions from the CAT Previous Year papers. In this article, we will look into some very important Logarithm questions PDF(with solutions) for CAT. You can also download these CAT Logarithms questions with detailed solutions, which also include important tricks to solve these questions.

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

Solution:

$2 log (2^x – 5) = log 2 + log (2^x – 7/2)$
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

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

Solution:

$x^u = 256$

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.

Question 3: 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)

Solution:

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

Question 4: 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

Solution:

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

Question 5: 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

Solution:

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

Question 6: 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

Solution:

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, $\log_6 G = a$

Question 7: 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

Solution:

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

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

Hence,

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

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

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

$– \frac{1}{6} + 1$

= $\frac{5}{6}$

Question 8: 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

Solution:

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

Question 9: 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

Solution:

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

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

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

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

Question 10: 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) $1+\log_{5}({\frac{3}{5}})$

Solution:

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.

Question 11: 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) $\log_{6}{8}$

Solution:

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.

Question 12: 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

Solution:

$\log_{2}({5+\log_{3}{a}})=3$
=>$5 + \log_{3}{a}$ = 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.

Question 13: $\frac{1}{log_{2}100}-\frac{1}{log_{4}100}+\frac{1}{log_{5}100}-\frac{1}{log_{10}100}+\frac{1}{log_{20}100}-\frac{1}{log_{25}100}+\frac{1}{log_{50}100}$=?

a) $\frac{1}{2}$

b) 10

c) 0

d) −4

Solution:

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 $log_{10}{100}=2$

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

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

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

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

Question 14: 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

Solution:

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

Question 15: 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) $-\log_{2}(\frac{1}{3})$

$2^{Y^2({\log_{3}{5})}}=5^{Y^2(\log_3 2)}$
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=$\left(-\log_23\right)=\left(\log_2\frac{1}{3}\right)$