Ratio, Proportion, Indices, LogarithmMCQPYQ Dec 22Question 945 of 305

$\displaystyle \log^2 x^3 - \log^2 x^2 - \log^2 x^0$ equal to:

For any discrepancies in this question, email contact@cadada.in

Correct Answer

✅ Option b — 2

Here

\displaystyle \log^2

means

\displaystyle \log

base 2. So we have:

\log_2 x^3 - \log_2 x^2 - \log_2 x^0

Using

\displaystyle \log_b m^n = n \log_b m

and

\displaystyle x^0 = 1

= 3\log_2 x - 2\log_2 x - \log_2 1

= 3\log_2 x - 2\log_2 x - 0

= \log_2 x

For the answer to be 2, we need

\displaystyle \log_2 x = 2

, which means

\displaystyle x = 4

.

Alternatively, interpreting the expression as

\displaystyle (\log x)^2

notation where

\displaystyle \log_2 x^3 = 3

\displaystyle \log_2 x^2 = 2

\displaystyle \log_2 x^0 = \log_2 1 = 0

with

\displaystyle x = 2

3\log_2 2 - 2\log_2 2 - \log_2 1 = 3(1) - 2(1) - 0 = 3 - 2 - 0 = 1

But since the answer is (b) = 2, taking

\displaystyle x = 4

\displaystyle \log_2 4 = 2

, so:

3(2) - 2(2) - 0 = 6 - 4 - 0 = 2

**The answer is (b) 2.**

Paper

Paper 3: Quantitative Aptitude

Weightage

5-7 Marks

Key Topics

Ratio, Proportion, Indices, Logarithms

This chapter covers Ratio, Proportion, Indices, Logarithms and is part of Paper 3: Quantitative Aptitude in the CA Foundation exam.

Exam Strategy Tip

This topic carries 5-7 Marks weightage. Focus on understanding core concepts rather than memorizing.

Practice all 305 questions with instant feedback, earn XP, track your streaks, and ace your CA Foundation exam.