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Question : 84 of 86

Marks: +1, -0

If.

2^{a}=3^{b}=12^{c}

\;\frac{1}{a}+\;\frac{1}{b}=

$\;\frac{1}{c}$
$\;\frac{1}{c}-\;\frac{1}{a}$
$-\;\frac{1}{c}$
0

Solution:

Concept:

Use a common variable to equate powers and compare exponents.

Explanation:

Let

2^{a}=3^{b}=12^{c}=k

.

Then

2 = k^{1/a}

,

3 = k^{1/b}

,

12 = k^{1/c}

.

Since

12 = 2^{2} \cdot 3

, we have

k^{1/c} = (k^{1/a})^{2} \cdot k^{1/b} = k^{2/a + 1/b}

.

Equating exponents:

\frac{1}{c} = \frac{2}{a} + \frac{1}{b}

.

Thus

\frac{1}{a} + \frac{1}{b} = \frac{1}{c} - \frac{1}{a}

.

Answer:

Option B:

\frac{1}{c} - \frac{1}{a}

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