GATE 2024 EC GA Q.4: solution and answer

GATE 2024 · ECGA Q.4 · 1 mark · MCQ

For a real number $x > 1$ ,

\frac{1}{lo g _{2} x} + \frac{1}{lo g _{3} x} + \frac{1}{lo g _{4} x} = 1

The value of $x$ is

(A)
4
(B)
12
(C)
24
(D)
36

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Answer and solution

ANSWER

Using $\frac{1}{lo g _{b} x} = lo g_{x} b$ , the equation becomes $lo g_{x} 2 + lo g_{x} 3 + lo g_{x} 4 = lo g_{x} 24 = 1$ , so $x = 24$ . The answer is option (C).

Official answer: CSource: Official answer key, GATE 2024 EC (IISc Bengaluru)

Reciprocal (base-swap) rule for logarithms: $\frac{1}{lo g _{b} x} = lo g_{x} b$ . It applies here because $x > 1$ , so $lo g_{x}$ is defined (base positive and not 1) and each $lo g_{b} x \neq = 0$ for $b = 2, 3, 4$ .
The equation therefore becomes $lo g_{x} 2 + lo g_{x} 3 + lo g_{x} 4 = 1$
Product rule, since all three terms now have the same base $x$ : $lo g_{x} a + lo g_{x} b + lo g_{x} c = lo g_{x} (ab c)$ . The left side is $lo g_{x} (2 \times 3 \times 4) = lo g_{x} 24$ .
$lo g_{x} 24 = 1$ means $x^{1} = 24$ , so $x = 24$ . This satisfies $x > 1$ .
Check: $lo g_{24} 2 + lo g_{24} 3 + lo g_{24} 4 = lo g_{24} (2 \times 3 \times 4) = lo g_{24} 24 = 1$ . The answer is option (C).

KEY IDEA

Swap the base with $\frac{1}{lo g _{b} x} = lo g_{x} b$ . All three terms then share the base $x$ , so they add into one logarithm of the product: $lo g_{x} (2 \cdot 3 \cdot 4) = 1$ , which gives $x = 2 \cdot 3 \cdot 4$ .

WATCH OUT FOR

Adding the bases ( $2 + 3 + 4 = 9$ , not an option) or taking their LCM (12, option B) when a sum of logarithms with the same base combines into the log of the PRODUCT, 24. Another slip is reading $\frac{1}{lo g _{b} x}$ as $lo g_{b} \frac{1}{x} = - lo g_{b} x$ . That identity is false: the reciprocal of a logarithm swaps the base and the argument, it does not invert the argument.

From the GATE 2024 EC paper. Every question of this paper · Download the full paper · All 58 Numerical questions, with notes

GATE 2024 EC GA Q.4: solution and answer

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