Ratio, Proportion, Indices, LogarithmMCQPYQ Dec. 22Question 886 of 305

If $\displaystyle (\sqrt{3})^{18} = (\sqrt{9})^x$ , find $\displaystyle x$ ?

A18

D19

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

Correct Answer

✅ Option b — 9

Given the equation:

(\sqrt{3})^{18} = (\sqrt{9})^x

Let us simplify both sides of the equation by writing them with base

\displaystyle 3

:
- Left-hand side (LHS):

\text{LHS} = (\sqrt{3})^{18} = \left(3^{\frac{1}{2}}\right)^{18} = 3^{\frac{18}{2}} = 3^9

- Right-hand side (RHS):
Since

\displaystyle \sqrt{9} = 3

, we have:

\text{RHS} = (\sqrt{9})^x = (3)^x = 3^x

Now equate the LHS and RHS:

3^9 = 3^x

Since the bases are identical, we equate the exponents:

x = 9

This matches **Option B**.
Note: The textbook answer key incorrectly specifies **Option C** (

\displaystyle 8

) as correct. However, our rigorous mathematical derivation clearly proves that

\displaystyle x = 9

(Option B).
Hence, **Option B** is the correct answer.

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.