Ratio, Proportion, Indices, LogarithmsPYQ May 25Question 4302 of 220
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If log⁡(a+b4)=12(log⁡a+log⁡b)\displaystyle \log \left(\frac{a+b}{4}\right) = \frac{1}{2}(\log a + \log b), then the value of ab+ba\displaystyle \frac{a}{b} + \frac{b}{a} will be

Options

A12
B14
C16
D8
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Correct Answer

✅ Option b — 14

All Options:

  • A12
  • B14
  • C16
  • D8

Detailed Solution & Explanation

We are given the logarithmic equation:
log⁡(a+b4)=12(log⁡a+log⁡b)\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)\log a + \log b = \log(ab)
Substituting this back into the equation, we get:
log⁡(a+b4)=12log⁡(ab)\log \left(\frac{a+b}{4}\right) = \frac{1}{2} \log(ab)
Using the power rule of logarithms, plog⁡q=log⁡(qp)\displaystyle p \log q = \log(q^p):
log⁡(a+b4)=log⁡((ab)1/2)=log⁡ab\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:
a+b4=ab\frac{a+b}{4} = \sqrt{ab}
Now, square both sides to remove the square root:
(a+b4)2=(ab)2\left(\frac{a+b}{4}\right)^2 = (\sqrt{ab})^2
(a+b)216=ab\frac{(a+b)^2}{16} = ab
(a+b)2=16ab(a+b)^2 = 16ab
Expand the left-hand side:
a2+2ab+b2=16aba^2 + 2ab + b^2 = 16ab
Subtract 2ab\displaystyle 2ab from both sides:
a2+b2=16ab−2aba^2 + b^2 = 16ab - 2ab
a2+b2=14aba^2 + b^2 = 14ab
To find the value of ab+ba\displaystyle \frac{a}{b} + \frac{b}{a}, we divide both sides by ab\displaystyle ab:
a2+b2ab=14abab\frac{a^2 + b^2}{ab} = \frac{14ab}{ab}
a2ab+b2ab=14\frac{a^2}{ab} + \frac{b^2}{ab} = 14
ab+ba=14\frac{a}{b} + \frac{b}{a} = 14
Hence, **Option B** is the correct answer.

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