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Question

Let (x0,y0) be the solution of the following equations (2x)ln2=(3y)ln3 , 3lnx=2lny then x0 is

A
16
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B
13
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C
12
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D
6
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Solution

The correct option is C 12
We have,
(2x)ln2=(3y)ln3ln2.ln2x=ln3.ln3yln2.ln2x=ln3.(ln3+lny)....(1)
Also given 3lnx=2lny
lnx.ln3=lny.ln2lny=lnx.ln3ln2
Substituting this value of lny in equation (1), we get
ln2.ln2x=ln3[ln3+lnx.ln3ln2](ln2)2ln2x=(ln3)2ln2+(ln3)2lnx(ln2)2ln2x=(ln3)2(ln2+lnx)(ln2)2ln2x(ln3)2ln2x=0[(ln2)2(ln3)2]ln2x=0ln2x=02x=1 or x=12

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