GATE 2026 EC GA Q.4: solution and answer

GATE 2026 · ECGA Q.4 · 1 mark · MCQ

Real numbers $y$ , $p$ , and $n$ (all greater than 1) satisfy

(lo g_{p^{1/ n}} y) (lo g_{y^{1/ n}} p) = 16,

where the logarithms are taken to the bases $p^{1/ n}$ and $y^{1/ n}$ .

The value of $n$ is ______

(A)
2
(B)
4
(C)
8
(D)
16

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Answer and solution

ANSWER

$n = 4$ , option (B).

Official answer: BSource: Official answer key, GATE 2026 EC (IIT Guwahati)

Power-of-base rule: $lo g_{a^{k}} x = \frac{ln x}{k ln a} = \frac{1}{k} lo g_{a} x$ . With $k = \frac{1}{n}$ this gives $lo g_{a^{1/ n}} x = n lo g_{a} x$ .
So $lo g_{p^{1/ n}} y = n lo g_{p} y$ and $lo g_{y^{1/ n}} p = n lo g_{y} p$ .
Multiply the two: $(lo g_{p^{1/ n}} y) (lo g_{y^{1/ n}} p) = n^{2} (lo g_{p} y) (lo g_{y} p)$
Reciprocal rule: $lo g_{p} y \cdot lo g_{y} p = \frac{ln y}{ln p} \cdot \frac{ln p}{ln y} = 1$ . This is valid because $p, y > 1$ , so $ln p \neq = 0$ and $ln y \neq = 0$ : both bases are valid and neither log is zero.
The equation becomes $n^{2} = 16$ , so $n = \pm 4$ .
The question says $n > 1$ , so $n = 4$ . This is option (B).

KEY IDEA

Taking the $n$ th root of a log's base multiplies the log by $n$ : $lo g_{a^{1/ n}} x = n lo g_{a} x$ . Each factor therefore gains a factor of $n$ , and $lo g_{p} y \cdot lo g_{y} p = 1$ . The whole product is $n^{2}$ , so $n^{2} = 16$ .

WATCH OUT FOR

Treating $lo g_{p^{1/ n}} y$ as $\frac{1}{n} lo g_{p} y$ (a root in the base multiplies the log by $n$ , it does not divide it). That gives $\frac{1}{n ^{2}} = 16$ , which no option satisfies. The other common slip is stopping at $n = 16$ and forgetting that each of the two factors contributes an $n$ , so the product holds $n^{2}$ , not $n$ .

From the GATE 2026 EC paper. Every question of this paper · Download the full paper · All 58 Numerical questions, with notes

GATE 2026 EC GA Q.4: solution and answer

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