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← IIT-JEE 2011 · Paper 1
MathematicsSingle correct+3 / -1Question 47 of 69

IIT-JEE 2011 · Paper 1, Mathematics question 47

Let (x0,y0)(x_0,y_0) be the solution of the following equations
(2x)ln⁡2=(3y)ln⁡3,3ln⁡x=2ln⁡y.(2x)^{\ln2}=(3y)^{\ln3},\qquad3^{\ln x}=2^{\ln y}.
Then x0x_0 is

Choose one option
Answer and worked solution

Official answer: C

Take logs.
ln⁡2(ln⁡2+ln⁡x)=ln⁡3(ln⁡3+ln⁡y),ln⁡3 ln⁡x=ln⁡2 ln⁡y.\ln2(\ln2+\ln x)=\ln3(\ln3+\ln y),\qquad\ln3\,\ln x=\ln2\,\ln y.

Substitute ln⁡y=ln⁡3ln⁡2ln⁡x\ln y=\frac{\ln3}{\ln2}\ln x into the first:
ln⁡x(ln⁡2−ln⁡23ln⁡2)=ln⁡23−ln⁡22  ⟹  ln⁡x⋅ln⁡22−ln⁡23ln⁡2=−(ln⁡22−ln⁡23),\ln x\left(\ln2-\frac{\ln^23}{\ln2}\right)=\ln^23-\ln^22\implies\ln x\cdot\frac{\ln^22-\ln^23}{\ln2}=-(\ln^22-\ln^23),
so ln⁡x=−ln⁡2\ln x=-\ln2 and x0=12x_0=\frac12.

(C)

Topics:Logarithms
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