JEE Advanced 2024 Paper 1 · Q08 · Sequences & Series

Question

Let $a = 3\sqrt{2}$ and $b = \dfrac{1}{5^{1/6}\sqrt{6}}$. If $x, y \in \mathbb{R}$ are such that $$3x + 2y = \log_a (18)^{5/4}\quad	ext{and}\quad 2x - y = \log_b\!\left(\sqrt{1080}ight),$$ then $4x + 5y$ is equal to ______.

SolveKar AI · Accepted Answer

Compute $\log_a(18)^{5/4}=\dfrac{5}{4}\log_{3\sqrt{2}}18$. $3\sqrt{2}=\sqrt{18}$, so $\log_{\sqrt{18}}18=2$. Hence $3x+2y=\dfrac{5}{4}\cdot 2=\dfrac{5}{2}$. Compute $\log_b\sqrt{1080}=\dfrac{1}{2}\log_b 1080$. $b=5^{-1/6}\cdot 6^{-1/2}$, so $\log_b N=\dfrac{\ln N}{-\frac{1}{6}\ln 5-\frac{1}{2}\ln 6}$. Note $1080=2^3\cdot 3^3\cdot 5=216\cdot 5=6^3\cdot 5$, so $\ln 1080=3\ln 6+\ln 5$. Denominator: $-\frac{1}{6}\ln 5-\frac{1}{2}\ln 6=-\frac{\ln 5+3\ln 6}{6}$. Therefore $\log_b 1080=-6$, and $\log_b\sqrt{1080}=-3$. So $2x-y=-3$. Solve: from $2x-y=-3\Rightarrow y=2x+3$. Substitute: $3x+2(2x+3)=\frac{5}{2}\Rightarrow 7x+6=\frac{5}{2}\Rightarrow 7x=-\frac{7}{2}\Rightarrow x=-\frac{1}{2}$, $y=2$. Then $4x+5y=-2+10=8$.

JEE Advanced 2024 Paper 1 · Q08 · Sequences & Series

Solution