GATE 2025 EE GA Q.5: solution and answer

GATE 2025 · EEGA Q.5 · 1 mark · MCQ

What is the value of $(\frac{3 ^{81}}{2 7 ^{4}})^{\frac{1}{3}}$ ?

(A)
$3^{13}$
(B)
$3^{96}$
(C)
$3^{23}$
(D)
$3^{69}$

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Answer and solution

ANSWER

$(\frac{3 ^{81}}{2 7 ^{4}})^{\frac{1}{3}} = (3^{69})^{\frac{1}{3}} = 3^{23}$ , so the answer is option (C).

Official answer: CSource: Official answer key, GATE 2025 EE (IIT Roorkee)

Write everything in base 3. Since $27 = 3^{3}$ , the rule $(a^{m})^{n} = a^{mn}$ gives $2 7^{4} = (3^{3})^{4} = 3^{12}$ .
To divide powers of the same base, subtract the exponents ( $a^{m} / a^{n} = a^{m - n}$ ): $\frac{3 ^{81}}{3 ^{12}} = 3^{81 - 12} = 3^{69}$ .
To raise this to the power $\frac{1}{3}$ , multiply the exponents: $(3^{69})^{\frac{1}{3}} = 3^{69/3} = 3^{23}$ .
Check by taking the cube root first: $(3^{81})^{\frac{1}{3}} = 3^{27}$ and $(2 7^{4})^{\frac{1}{3}} = (3^{12})^{\frac{1}{3}} = 3^{4}$ , so the value is $\frac{3 ^{27}}{3 ^{4}} = 3^{23}$ again.
The value is $3^{23}$ , which is option (C).

KEY IDEA

Write every term in the same base (here $27 = 3^{3}$ ) and then work only with exponents: multiply for a power of a power, and subtract for a quotient.

WATCH OUT FOR

Stopping at $3^{81 - 12} = 3^{69}$ and forgetting the outer power $\frac{1}{3}$ , which gives option (D). The other slip is to leave $2 7^{4}$ as it is, or to write it as $3^{7}$ by adding $3 + 4$ when the exponents should be multiplied to give $3^{12}$ .

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GATE 2025 EE GA Q.5: solution and answer

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