Basics & LogarithmsmediumPYQ 2011 · JEE Advanced · Paper 1Free

Basics & Logarithms: Let Solution Following Equations (JEE Advanced 2011)

JEE Maths question with a full step-by-step solution.

Question
Let (x0,y0)(x_{0},y_{0}) be the solution of the following equations: (2x)ln⁡2=(3y)ln⁡3(2x)^{\ln 2}=(3y)^{\ln 3} and 3ln⁡x=2ln⁡y3^{\ln x}=2^{\ln y}. Then x0x_{0} is
A16\dfrac{1}{6}
B13\dfrac{1}{3}
C12\dfrac{1}{2}correct
D66
Solution
Step 1: Take ln⁡\ln of both given equations. From (2x)ln⁡2=(3y)ln⁡3(2x)^{\ln 2}=(3y)^{\ln 3}:
ln⁡2⋅(ln⁡2+ln⁡x)=ln⁡3⋅(ln⁡3+ln⁡y)\ln 2\cdot(\ln 2+\ln x)=\ln 3\cdot(\ln 3+\ln y)
(ln⁡2)2+ln⁡2⋅ln⁡x=(ln⁡3)2+ln⁡3⋅ln⁡y…(1)(\ln 2)^{2}+\ln 2\cdot\ln x=(\ln 3)^{2}+\ln 3\cdot\ln y\quad\ldots(1)
 Step 2: From 3ln⁡x=2ln⁡y3^{\ln x}=2^{\ln y}:
ln⁡x⋅ln⁡3=ln⁡y⋅ln⁡2  ⇒  ln⁡y=ln⁡x⋅ln⁡3ln⁡2…(2)\ln x\cdot\ln 3=\ln y\cdot\ln 2\;\Rightarrow\;\ln y=\dfrac{\ln x\cdot\ln 3}{\ln 2}\quad\ldots(2)
 Step 3: Substitute (2) into (1):
(ln⁡2)2+ln⁡2⋅ln⁡x=(ln⁡3)2+ln⁡3⋅ln⁡x⋅ln⁡3ln⁡2(\ln 2)^{2}+\ln 2\cdot\ln x=(\ln 3)^{2}+\ln 3\cdot\dfrac{\ln x\cdot\ln 3}{\ln 2}
(ln⁡2)2+ln⁡2⋅ln⁡x=(ln⁡3)2+(ln⁡3)2⋅ln⁡xln⁡2(\ln 2)^{2}+\ln 2\cdot\ln x=(\ln 3)^{2}+\dfrac{(\ln 3)^{2}\cdot\ln x}{\ln 2}
 Step 4: Rearrange:
(ln⁡2)2−(ln⁡3)2=(ln⁡3)2⋅ln⁡xln⁡2−ln⁡2⋅ln⁡x(\ln 2)^{2}-(\ln 3)^{2}=\dfrac{(\ln 3)^{2}\cdot\ln x}{\ln 2}-\ln 2\cdot\ln x
(ln⁡2)2−(ln⁡3)2=ln⁡x((ln⁡3)2ln⁡2−ln⁡2)(\ln 2)^{2}-(\ln 3)^{2}=\ln x\left(\dfrac{(\ln 3)^{2}}{\ln 2}-\ln 2\right)
(ln⁡2)2−(ln⁡3)2=ln⁡x⋅(ln⁡3)2−(ln⁡2)2ln⁡2(\ln 2)^{2}-(\ln 3)^{2}=\ln x\cdot\dfrac{(\ln 3)^{2}-(\ln 2)^{2}}{\ln 2}
 Step 5: Simplify:
−((ln⁡3)2−(ln⁡2)2)=ln⁡x⋅(ln⁡3)2−(ln⁡2)2ln⁡2-((\ln 3)^{2}-(\ln 2)^{2})=\ln x\cdot\dfrac{(\ln 3)^{2}-(\ln 2)^{2}}{\ln 2}
−1=ln⁡xln⁡2  ⇒  ln⁡x=−ln⁡2=ln⁡12-1=\dfrac{\ln x}{\ln 2}\;\Rightarrow\;\ln x=-\ln 2=\ln\dfrac{1}{2}
x0=12x_{0}=\dfrac{1}{2}
 Answer: (3) 12\dfrac{1}{2}
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