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Question

# Let (x0,y0) be thhe solution of the following equations (2x)ln2=(3y)ln3 and 3ln3 and 3lnx=2lny. Then xy=

A
1/6
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B
1/3
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C
1/2
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D
0
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Solution

## The correct option is D 1/6(2x)ln2=(3y)ln3Taking ln both side,⇒(ln2)ln(2x)=(ln3)ln(3y)⇒(ln2)(ln2+lnx)=(ln3)(ln3+lny)⇒(ln2)(lny)(ln2+lnx)=(ln3)(1+ln3lny)⇒(ln2)(ln2lny+lnxlny)=ln3(1+ln3lny)(equation 1)And 3lnx=2lnyTaking ln both side,⇒(lnx)(ln3)=(lny)(ln2)⇒lnxlny=ln2ln3(equation 2)From 1 and 2⇒(ln2)(ln2lny+lnxlny)=ln3(1+ln3lny)⇒(ln2)2−(ln3)2lny=(ln3)2(ln2)2ln3⇒lny=−ln3⇒y=13And from 1 ⇒lnx=−ln2⇒x=+12

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