CAT Logarithms Questions & Solutions

A sample of real CAT Logarithms past-year questions, each with a worked solution and the correct answer highlighted. AthenaPrep has 54 Logarithms questions in all — sign up free to practise them timed.

1. Q1.CAT 2025

   The number of distinct integers $n$ for which \log_{\frac{1}{4}}({n^{2}-7n+11})>0,is

   • 2

   • infinite

   • 1

   • 0

   Show solution

   For base of log in range  $1/4\in(0,1)$ and \log_{1/4}(x)&gt;0 is true only if 0<x<1. 

   For integer n, $x=n^2-7n+11$ is an integer, so it cannot lie strictly between 0 and 1.

   So, there is no integer value for which this inequality is satisfied.
   

2. Q2.CAT 2024

   If $(x + 6\sqrt{2})^{\cfrac{1}{2}} - (x - 6\sqrt{2})^{\cfrac{1}{2}} = 2\sqrt{2}$, then x equals

   Answer: 11

   Show solution

   Squaring on both sides, we get:

   $x+6\sqrt{\ 2}+x-6\sqrt{\ 2}-2\left(x^2-72\right)^{\frac{1}{2}}=8$
   $x-\left(x^2-72\right)^{\frac{1}{2}}=4$
   
   Bringing x to the other side, we get:
   $-\left(x^2-72\right)^{\frac{1}{2}}=4-x$
   Squaring on both sides again, we get:

   $x^2-72=16+x^2-8x$
   $8x=88$
   $x=11$

   Therefore, 11 is the correct answer. 

3. Q3.CAT 2024

   If $3^a = 4, 4^b = 5, 5^c = 6, 6^d = 7, 7^e = 8$ and $8^f = 9$, then the value of the product abcdef is

   Answer: 2

   Show solution

   Taking a log for each of the expressions, we get the following:

   $\log_34=a,\ \log_45=b,\ \log_56=c,\ \log_67=d,\ \log_78=e,\ \log_89=f$

   The expression $abcef$ would then be:  $\log_34\times\ \log_45\times\ \log_56\times\ \log_67\times\ \log_78\times\ \log_89$
   

   Next, we can use this property of log:  $\frac{\log_ba}{\log_bc}=\log_ca$
   Using this, we get:

   $\frac{\log\ 4}{\log\ 3}\times\ \frac{\log\ 5}{\log\ 4}\times\ \frac{\log\ 6}{\log\ 5}\times\ \frac{\log\ 7}{\log\ 6}\times\ \frac{\log\ 8}{\log\ 7}\times\ \frac{\log\ 9}{\log\ 8}$

   All the terms will cancel out except:  $\frac{\log\ 9}{\log\ 3}=\log_39=2$
   

   Therefore, 2 is the correct answer. 

4. Q4.CAT 2023

   For some positive real number x, if $\log_{\sqrt{3}}{(x)}+\frac{\log_{x}{(25)}}{\log_{x}{(0.008)}}=\frac{16}{3}$, then the value of $\log_{3}({3x^{2}})$ is

   Answer: 7

   Show solution

   It is given that  $\log_{\sqrt{3}}{(x)}+\frac{\log_{x}{(25)}}{\log_{x}{(0.008)}}=\frac{16}{3}$ , which can be written as:

   => $2\log_3x+\log_{0.008}25\ =\ \frac{16}{3}$

   => $2\log_3x+\log_{\frac{8}{1000}}25\ =\ \frac{16}{3}$

   => $2\log_3x+\log_{\frac{1}{125}}25\ =\ \frac{16}{3}$

   => $2\log_3x+\log_{5^{-3}}\left(5\right)^2\ =\ \frac{16}{3}$

   => $2\log_3x-\frac{2}{3}=\ \frac{16}{3}$

   => $2\log_3x=\frac{16}{3}+\frac{2}{3}$

   => $2\log_3x=6$

   => \log_3x^2=6\ =&gt;\ x^2\ =\ 3^6

   Hence,  $\log_3\left(3\cdot x^2\right)\ =\ \log_3\left(3\cdot3^6\right)\ =\log_33^7\ =\ 7$

5. Q5.CAT 2020

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

   • $\log_{2}(\frac{1}{5})$

   • $\log_{2}(\frac{1}{3})$

   • $-\log_{2}(\frac{1}{5})$

   • $-\log_{2}(\frac{1}{3})$

   Show solution

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

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

   =>  Y^2\left(\log_32\right)=\left(\log_23\right)=&gt;Y^2=\left(\log_23\right)^2

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

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

6. Q6.CAT 2020

   If a,b,c are non-zero and $14^a=36^b=84^c$, then $6b(\frac{1}{c}-\frac{1}{a})$ is equal to

   Answer: 3

   Show solution

   Let  $14^a=36^b=84^c$ = k

   => a =  $\log_{14}k$ , b =  $\log_{36}k$ , c= $\log_{84}k$

   $6b(\frac{1}{c}-\frac{1}{a})$ =  $6\cdot\frac{1}{2}\log_6k\left(\log_k84-\log_k14\right)$ = 3

7. Q7.CAT 2019

   If m and n are integers such that $(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$ then m is

   • -20

   • -24

   • -12

   • -16

   Show solution

   We have,  $(\surd2)^{19} 3^4 4^2 9^m 8^n = 3^n 16^m (\sqrt[4]{64})$

   Converting both sides in powers of 2 and 3, we get

   $2^{\ \frac{19\ }{2}}3^42^43^{2m}2^{3n}$ =  $3^n2^{4m}2^{\frac{\ 6}{4}}$

   Comparing the power of 2 we get,  \ \frac{\ 19}{2}+4+3n\ =4m+\frac{\ 6}{4}\

   => 4m=3n+12 .....(1)

   Comparing the power of 3 we get,  $4+2m=n$

   Substituting the value of n in (1), we get

   4m=3(4+2m)+12

   => m=-12

8. Q8.CAT 2018

   Given that $x^{2018}y^{2017}=\frac{1}{2}$, and $x^{2016}y^{2019}=8$, then value of $x^{2}+y^{3}$ is

   • $\dfrac{31}{4}$

   • $\dfrac{35}{4}$

   • $\dfrac{37}{4}$

   • $\dfrac{33}{4}$

   Show solution

   Given that $x^{2018}y^{2017}=\frac{1}{2}$   ... (1)

   $x^{2016}y^{2019}=8$ ... (2)

   Equation (2)/ Equation (1)

   $\dfrac{y^2}{x^2} = \dfrac{8}{1/2}$

   $\dfrac{y}{x} = 4$ or $-4$
   

   Case 1: When  $\dfrac{y}{x} = 4$

   $x^{2018}(4x)^{2017}=\dfrac{1}{2}$
   

   $x^{2018+2017}(2)^{4034}=\dfrac{1}{2}$
   

   $x^{4035}=\dfrac{1}{(2)^{4035}}$
   

   $x=\dfrac{1}{2}$

   Since, $\dfrac{y}{x} = 4$ , => y = 2
   

   Therefore,  $x^{2}+y^{3}$ = $\dfrac{1}{4}+8$ = $\dfrac{33}{4}$

   Case 2: When  $\dfrac{y}{x} = -4$

   $x^{2018}(-4x)^{2017}=\dfrac{1}{2}$

   $x^{2018+2017}(2)^{4034}=\dfrac{-1}{2}$

   $x^{4035}=\dfrac{1}{(-2)^{4035}}$

   $x=\dfrac{-1}{2}$

   Since, $\dfrac{y}{x} = -4$ , => y = 2

   Therefore,  $x^{2}+y^{3}$ = $\dfrac{1}{4}+8$ =  $\dfrac{33}{4}$ . Hence, option D is the correct answer. 

9. Q9.CAT 2017

   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

   • $\sqrt{a}$

   • 2a

   • a/2

   • a

   Show 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, $\log_6 G = a$

10. Q10.CAT 2017

    If $9^{x-\frac{1}{2}}-2^{2x-2}=4^{x}-3^{2x-3}$, then $x$ is
    

    • 3/2

    • 2/5

    • 3/4

    • 4/9

    Show solution

    It is given that  $9^{x-\frac{1}{2}}-2^{2x-2}=4^{x}-3^{2x-3}$

    Let us try to reduce them to powers of $3$ and $2$
    The given equation can be reduced to $3^{2x-1} + 3^{2x-3} = 2^{2x} + 2^{2x-2}$
    

    Hence, $3^{2x-3} \times 10 = 2^{2x-2} \times 5$
    Therefore, $3^{2x-3} = 2^{2x-3}$

    This is possible only if $2x-3=0$ or $x=3/2$

11. Q11.CAT 2006

    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?

    • (-2, 1/2)

    • (1,1)

    • (0.4, 2.5)

    • ($\pi$, 1/ $\pi$)

    • (2,2)

    Show 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

12. Q12.CAT 2002

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

    • $f(x+y)$

    • $f{\frac{(x+y)}{(1+xy)}}$

    • $(x+y)f{\frac{1}{(1+xy)}}$

    • $\frac{f(x)+f(y)}{(1+xy)}$

    Show solution

    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.

CAT Logarithms previous year questions with solutions — AthenaPrep