JEE Advanced 2024 — Paper 1 — MATH-III solution

JEE Advanced2024Paper 1MATH-III

Let $a = 32$ and $b = \frac{1}{5 ^{1/6} 6}$ . If $x, y \in R$ are such that

$3 x + 2 y = lo g_{a} (18)^{5/4}$ and

$2 x - y = lo g_{b} (1080)$ ,

then $4 x + 5 y$ is equal to ________.

Your answer

Solution

Since $a = 32$ , we get $a^{2} = 18$ . Therefore

$3 x + 2 y = lo g_{a} (18)^{5/4} = lo g_{a} (a^{2})^{5/4} = \frac{5}{2} \dots (i)$

Note that $1080 = 5 \cdot 6^{3}$ , so $108 0^{1/6} = 5^{1/6} \cdot 6^{1/2} = \frac{1}{b}$ , hence $1080 = 108 0^{1/2} = b^{- 3}$ .

$2 x - y = lo g_{b} (b^{- 3}) = - 3 \dots (ii)$

Solving (i) and (ii) gives $x = - \frac{1}{2}$ , $y = 2$ .

$∴ 4 x + 5 y = - 2 + 10 = 8$

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