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Logarithms questions for IPMAT, with solutions

10 free Logarithms practice questions for IPMAT Indore, Rohtak and JIPMAT, 7 from past papers, each with the answer and a full worked solution. No sign-in. Try each one first, then open the solution. The lessons for this chapter are in Logarithms on Learn.

1. Question 1Type the answerWritten by IPMATwayLogarithms with Progressions

   If $\log_2 a$, $\log_2 b$ and $\log_2 c$ are in arithmetic progression with $a=4$ and $c=64$, find $b$.

   Show the answer and solution

   Answer: 16

   AP of logs means $2\log_2b=\log_2a+\log_2c$, i.e. $b^2=ac$. So $b^2=4\cdot64=256$. Since $b>0$ (it is inside a log), $b=16$. Check: the logs are $2,4,6$, an AP.

2. Question 2Type the answerWritten by IPMATwayLogarithms with Progressions

   In a geometric progression of positive terms the $2$nd term is $8$ and the $6$th term is $128$. Find $\log_2(T_1\cdot T_7)$, where $T_k$ is the $k$th term.

   Show the answer and solution

   Answer: 10

   Here $r^4=\frac{T_6}{T_2}=\frac{128}{8}=16$, so $r=2$ (positive terms). Then $T_1=\frac{8}{2}=4$ and $T_7=4\cdot2^6=256$. So $T_1T_7=1024=2^{10}$ and $\log_2(T_1T_7)=10$.

3. Question 3Multiple choiceAsked in IPMAT Indore 2024Log Laws and Evaluation

   If $\log_4 x=a$ and $\log_{25}x=b$ then $\log_x 10$ is

   1. A.$\frac{a+b}{2(a-b)}$
   2. B.$\frac{a+b}{2}$
   3. C.$\frac{a+b}{2ab}$
   4. D.$\frac{a-b}{2ab}$
   Show the answer and solution

   Answer: C. $\frac{a+b}{2ab}$

   Invert the given logs: $\log_x4=\frac1a$ and $\log_x25=\frac1b$. Halve each because $4=2^2$ and $25=5^2$. Then $\log_x10=\log_x2+\log_x5=\frac{b+a}{2ab}$.

4. Question 4Multiple choiceAsked in IPMAT Indore 2024Log Laws and Evaluation

   Let $a=\dfrac{(\log_7 4)(\log_7 5-\log_7 2)}{(\log_7 25)(\log_7 8-\log_7 4)}$. Then the value of $5^a$ is

   1. A.$\frac72$
   2. B.$5$
   3. C.$8$
   4. D.$\frac52$
   Show the answer and solution

   Answer: D. $\frac52$

   Simplify each factor: $\log_74=2\log_72$; $\log_75-\log_72=\log_7\frac52$; $\log_725=2\log_75$; $\log_78-\log_74=\log_72$. Cancel $2\log_72$: $a=\frac{\log_7(5/2)}{\log_75}=\log_5\frac52$. Hence $5^a=\frac52$.

5. Question 5Multiple choiceWritten by IPMATwayLog Laws and Evaluation

   If $a=1+\log_x(yz)$, $b=1+\log_y(zx)$ and $c=1+\log_z(xy)$, then $ab+bc+ca$ equals

   1. A.$a+b+c$
   2. B.$0$
   3. C.$1$
   4. D.$abc$
   Show the answer and solution

   Answer: D. $abc$

   $1+\log_x(yz)=\log_xx+\log_x(yz)=\log_x(xyz)$. Reciprocals: $\frac1a=\log_{xyz}x$, $\frac1b=\log_{xyz}y$, $\frac1c=\log_{xyz}z$, which add to $1$. Multiply $\frac1a+\frac1b+\frac1c=1$ by $abc$.

6. Question 6Multiple choiceAsked in IPMAT Indore 2023Logarithmic Equations

   If $\log_{\cos x}(\sin x)+\log_{\sin x}(\cos x)=2$, then the value of $x$ is

   1. A.$n\pi+\frac{\pi}{4}$, $n$ is an integer
   2. B.$2n\pi+\frac{\pi}{4}$, $n$ is an integer
   3. C.$\frac{n\pi}{4}+\frac{\pi}{4}$, $n$ is an integer
   4. D.$2n\pi-\frac{\pi}{4}$, $n$ is an integer
   Show the answer and solution

   Answer: B. $2n\pi+\frac{\pi}{4}$, $n$ is an integer

   Let $t=\log_{\cos x}\sin x$; the equation is $t+\frac1t=2$, so $t=1$, i.e. $\sin x=\cos x$. Bases must be positive and $\ne1$, so $\sin x=\cos x=\frac1{\sqrt2}>0$: first-quadrant angle $\frac\pi4$ plus full turns, $x=2n\pi+\frac\pi4$.

7. Question 7Multiple choiceAsked in IPMAT Indore 2025Logarithmic Equations

   If $\log_{25}[5\log_3(1+\log_3(1+2\log_2 x))]=\frac12$ then $x$ is

   1. A.$16$
   2. B.$2$
   3. C.$8$
   4. D.$4$
   Show the answer and solution

   Answer: A. $16$

   Unwrap step by step: the bracket equals $25^{1/2}=5$, so $\log_3(1+\log_3(1+2\log_2x))=1$, giving $1+\log_3(1+2\log_2x)=3$, $\log_3(1+2\log_2x)=2$, $1+2\log_2x=9$, $\log_2x=4$, $x=16$.

8. Question 8Multiple choiceAsked in IPMAT Indore 2019Logarithmic Inequalities

   The inequality $\log_2\left(\frac{3x-1}{2-x}\right)<1$ holds true for

   1. A.$x\in\left(\frac13,1\right)$
   2. B.$x\in\left(\frac13,2\right)$
   3. C.$x\in\left(0,\frac13\right)\cup(1,2)$
   4. D.$x\in(-\infty,1)$
   Show the answer and solution

   Answer: A. $x\in\left(\frac13,1\right)$

   Base $2>1$, so the inequality means $0<\frac{3x-1}{2-x}<2$. The fraction is positive for $\frac13<x<2$; on that interval $2-x>0$, so multiply: $3x-1<4-2x$, i.e. $x<1$. Intersection: $\frac13<x<1$.

9. Question 9Multiple choiceAsked in IPMAT Indore 2022Logarithmic Inequalities

   The set of real values of $x$ for which the inequality $\log_{27}8\le\log_3 x<9^{\frac{1}{\log_2 3}}$ holds is

   1. A.$[2,81)$
   2. B.$(2,27)$
   3. C.$[2,81]$
   4. D.$(2,27]$
   Show the answer and solution

   Answer: A. $[2,81)$

   $\log_{27}8=\log_{3^3}2^3=\log_32$. Also $\frac1{\log_23}=\log_32$, so $9^{\log_32}=3^{2\log_32}=4$. Base $3>1$: $\log_32\le\log_3x<4\iff2\le x<3^4=81$.

10. Question 10Multiple choiceAsked in IPMAT Indore 2023Logarithmic Equations

    Let $a,b,c$ be real numbers greater than $1$, and $n$ be a positive real number not equal to $1$. If $\log_n(\log_2 a)=1$, $\log_n(\log_2 b)=2$ and $\log_n(\log_2 c)=3$, then which of the following is true?

    1. A.$a^n+b^n=c^n$
    2. B.$(a^n+b)^n=ac$
    3. C.$a+b=c$
    4. D.$(b-a)^n=(c-b)$
    Show the answer and solution

    Answer: B. $(a^n+b)^n=ac$

    Unwrap: $\log_2a=n$, $\log_2b=n^2$, $\log_2c=n^3$. Then $a^n=2^{n^2}=b$, so $a^n+b=2b=2^{n^2+1}$ and $(a^n+b)^n=2^{n^3+n}$. Also $ac=2^{n+n^3}$. Quick check with $n=2$: $a=4,b=16,c=256$: $(16+16)^2=1024=4\cdot256$.