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← JEE Advanced 2024 · Paper 1
MathematicsInteger+4Question 8 of 51

JEE Advanced 2024 · Paper 1, Mathematics question 8

Let a=32a = 3\sqrt{2} and b=151/66b = \dfrac{1}{5^{1/6}\sqrt{6}}. If x,y∈Rx, y \in \mathbb{R} are such that

3x+2y=log⁡a(18)54and3x + 2y = \log_a (18)^{\frac{5}{4}} \quad\text{and}
2x−y=log⁡b(1080),2x - y = \log_b\left(\sqrt{1080}\right),

then 4x+5y4x + 5y is equal to _____.

Answer and worked solution

Official answer: 8

First equation. a=32=181/2a = 3\sqrt2 = 18^{1/2}, so log⁡a185/4=5/41/2=52\log_a 18^{5/4} = \dfrac{5/4}{1/2} = \dfrac52. Hence 3x+2y=523x + 2y = \dfrac52.

Second equation. 1080=23⋅33⋅51080 = 2^3\cdot3^3\cdot5, so 1080=63/2 51/2\sqrt{1080} = 6^{3/2}\,5^{1/2}. Also b=5−1/6 6−1/2b = 5^{-1/6}\,6^{-1/2}, so b−3=51/2 63/2=1080b^{-3} = 5^{1/2}\,6^{3/2} = \sqrt{1080}. Hence 2x−y=−32x - y = -3.

Solve. y=2x+3y = 2x + 3, so 3x+4x+6=523x + 4x + 6 = \frac52, giving x=−12x = -\frac12 and y=2y = 2.

4x+5y=−2+10=84x + 5y = -2 + 10 = 8.

Topics:Logarithms