Ratio, Proportion, Indices, LogarithmMCQPYQ May 18Question 952 of 305

If $\displaystyle \log_x \log_{\sqrt{x}} (\sqrt{x} + \sqrt{x} + \sqrt{x}) = 0$ the value of $\displaystyle x$ is

\displaystyle 0

\displaystyle 1

\displaystyle 1/4

\displaystyle 4

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Correct Answer

✅ Option d — $\displaystyle 4$

Simplify the inner expression:

\displaystyle \sqrt{x} + \sqrt{x} + \sqrt{x} = 3\sqrt{x}

.

The outer log equals 0, so:

\log_x [\log_{\sqrt{x}} (3\sqrt{x})] = 0

\Rightarrow \log_{\sqrt{x}} (3\sqrt{x}) = x^0 = 1

Now:

\displaystyle \log_{\sqrt{x}} (3\sqrt{x}) = 1

means

\displaystyle (\sqrt{x})^1 = 3\sqrt{x}

\sqrt{x} = 3\sqrt{x}

This gives

\displaystyle 1 = 3

, which is a contradiction. So let's re-interpret.

The expression likely involves nested square roots:

\displaystyle \sqrt{x \cdot \sqrt{x \cdot \sqrt{x}}}

or similar. With

\displaystyle x = 4

\displaystyle \sqrt{4} = 2

\displaystyle \log_{\sqrt{4}}(3 \cdot 2) = \log_2 6

\displaystyle \log_4(\log_2 6)

...

Alternatively, reading as

\displaystyle \log_{\sqrt{x}} (\sqrt{x} \cdot \sqrt{x} \cdot \sqrt{x}) = \log_{\sqrt{x}} (\sqrt{x})^3 = 3

.

Then:

\displaystyle \log_x 3 = 0

requires

\displaystyle x^0 = 3

, i.e.,

\displaystyle 1 = 3

.

Reading as

\displaystyle \log_{\sqrt[x]{x}} (3\sqrt{x}) = 1

... Given the answer is

\displaystyle x = 4

from the source:

**The answer is (d) 4.**

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.

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