Logarithm Questions with Solutions

15 questions3 easy8 medium4 hard

A logarithm answers the question: to what power must a base be raised to give a number? If a^x = N, then log base a of N = x. This topic covers the laws of logarithms (product, quotient and power rules), change of base, common and natural logarithms, and solving logarithmic and exponential equations. Most exam questions reduce to applying the three main rules correctly: log(mn) = log m + log n, log(m/n) = log m - log n, and log(m^k) = k log m. Remember that logarithms are only defined for positive numbers and a base that is positive and not equal to 1, which helps you reject invalid answers. Practise the logarithm questions below. Each comes with a step-by-step solution showing which law is applied at each step.

All Questions Easy Medium Hard

Q1.
If $l o g_{3} 2 = x$ and $l o g_{3} 5 = y$ , find $l o g_{3} 1620$ .
A. $2 (x + 2) + y$ B. $2 x + y + 2$ C. $2 (x + y) + 1$ D. None of these
mediumAsked in MAH MCA CET 2023View solution ›
Q2.
If $l o g_{5} 2 = x$ , then find $l o g_{25} 800$ .
A. $2.5 x + 1$ B. $2.5 x + 2$ C. $1.5 x + 1$ D. $1.5 x + 3$
mediumAsked in MAH MCA CET 2023View solution ›
Q3.
The logarithm of the difference between the squares of two positive integers $A$ and $B$ is a fixed constant $C$ . If $A = B$ , then the value of $C$ is
A. None of theseB. 1.303C. $l o g 2$ D. $e^{2}$
mediumAsked in MAH MCA CET 2023View solution ›
Q4.
Which of the following numbers is the coefficient of $x^{100}$ in the expansion of $l o g_{e} (\frac{1 + x}{1 + x ^{2}})$ , $∣ x ∣ < 1$ ?
A. 0.01B. 0.02C. -0.03D. -0.01
mediumAsked in NIMCET 2023View solution ›
Q5.
The solution set of $l o g_{3} ((x + 2) (x + 4)) + l o g_{1/3} (x + 2) < \frac{1}{2} l o g_{3} 7$ is
A. $(- 2, - 1)$ B. $(- 2, 3)$ C. $(- 1, 3)$ D. $(3, \infty)$
mediumAsked in NIMCET 2011View solution ›
Q6.
If $l o g_{3} 2$ , $l o g_{3} (2^{x} - 5)$ , $l o g_{3} (2^{x} - \frac{7}{2})$ are in arithmetic progression, then the value of x is equal to
A. 5B. 4C. 3D. 2
mediumAsked in NIMCET 2018View solution ›
Q7.
If $8^{x - 1} = (\frac{1}{4})^{x}$ , then the value of $\frac{1}{l o g _{x + 1} 4 - l o g _{x + 1} 5} + \frac{1}{l o g _{1 - x} 4 - l o g _{1 - x} 5}$ is
A. 1B. $\frac{5}{4}$ C. $\frac{4}{5}$ D. 2
mediumAsked in NIMCET 2025View solution ›
Q8.
Let $x$ be a positive real number such that $x^{8 l o g_{5} x - 24} = 5^{- 4}$ . Then the product of all possible values of $x$ is
A. 5B. 125C. 165D. 25
mediumAsked in NIMCET 2025View solution ›