Ratio, Proportion, Indices, LogarithmMCQPYQ Nov. 18Question 931 of 305

$\displaystyle \log_x \log_x \log_x 16 = ?$

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Correct Answer

✅ Option c — 1

We are given the expression:

E = \log_x \log_x \log_x 16

Let us find the value of the expression for the standard base

\displaystyle x = 2

:
1) Evaluate the innermost logarithm:

\log_2 16 = \log_2\left( 2^4 \right) = 4 \log_2 2 = 4

2) Substitute this into the next logarithm:

\log_2 \left( \log_2 16 \right) = \log_2 4 = \log_2\left( 2^2 \right) = 2 \log_2 2 = 2

3) Substitute this into the outermost logarithm:

E = \log_2 \left( \log_2 \log_2 16 \right) = \log_2 2 = 1

Thus, under the standard base

\displaystyle x = 2

, the expression simplifies exactly to

\displaystyle 1

, which corresponds to Option C. If we solve the equation

\displaystyle \log_x \log_x \log_x 16 = 1

, we find:

\log_x \log_x 16 = x

\log_x 16 = x^x \implies 16 = x^{(x^x)}

By inspection, if

\displaystyle x = 2

\displaystyle 2^{(2^2)} = 2^4 = 16

, confirming that

\displaystyle x=2

is the unique integer solution that makes the expression equal to

\displaystyle 1

.

Hence, **Option C** 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.

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