Ratio, Proportion, Indices, LogarithmMCQPYQ May 25Question 4002 of 305

If $\displaystyle \log \left(\frac{a+b}{4}\right) = \frac{1}{2}(\log a + \log b)$ , then the value of $\displaystyle \frac{a}{b} + \frac{b}{a}$ will be

A12

B14

C16

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Correct Answer

✅ Option b — 14

We are given the logarithmic equation:

\log \left(\frac{a+b}{4}\right) = \frac{1}{2}(\log a + \log b)

Using the properties of logarithms, we know that:

\log a + \log b = \log(ab)

Substituting this back into the equation, we get:

\log \left(\frac{a+b}{4}\right) = \frac{1}{2} \log(ab)

Using the power rule of logarithms,

\displaystyle p \log q = \log(q^p)

\log \left(\frac{a+b}{4}\right) = \log\left((ab)^{1/2}\right) = \log\sqrt{ab}

By taking the antilogarithm on both sides, we eliminate the logarithm:

\frac{a+b}{4} = \sqrt{ab}

Now, square both sides to remove the square root:

\left(\frac{a+b}{4}\right)^2 = (\sqrt{ab})^2

\frac{(a+b)^2}{16} = ab

(a+b)^2 = 16ab

Expand the left-hand side:

a^2 + 2ab + b^2 = 16ab

Subtract

\displaystyle 2ab

from both sides:

a^2 + b^2 = 16ab - 2ab

a^2 + b^2 = 14ab

To find the value of

\displaystyle \frac{a}{b} + \frac{b}{a}

, we divide both sides by

\displaystyle ab

\frac{a^2 + b^2}{ab} = \frac{14ab}{ab}

\frac{a^2}{ab} + \frac{b^2}{ab} = 14

\frac{a}{b} + \frac{b}{a} = 14

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.

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