# Logarithm question-I donot know but this question may be solved by any other way also.Let (x_0,y_0) be the solution of the following equations. (2x)^(ln2)=(3y)^(ln3) 3^(lnx)=2^(lny) Then x_0 is A) 1/6 B) 1/3 C) 1/2 D) 6 I have tried this problem by taking log on both sides of the two equations. But, finally I could not make up to get the values of x and y.

Logarithm question-I donot know but this question may be solved by any other way also.
Let $\left({x}_{0},{y}_{0}\right)$ be the solution of the following equations.
$\left(2x{\right)}^{\mathrm{ln}2}=\left(3y{\right)}^{\mathrm{ln}3}$
${3}^{\mathrm{ln}x}={2}^{\mathrm{ln}y}$
Then ${x}_{0}$ is
A) $\frac{1}{6}$
B) $\frac{1}{3}$
C) $\frac{1}{2}$
D) $6$
I have tried this problem by taking log on both sides of the two equations. But, finally I could not make up to get the values of $x$ and $y$
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wintern90
I would suggest to:
take $\mathrm{ln}$ on both sides of the first equation.
express $\mathrm{ln}y$ from the second equation and substitute it into the first equation
Now you have the equation with one variable. After rather simple transformations you will get the answer for $\mathrm{ln}x$ and later for $x$
Let me know if I'm not clear or you need further help.
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kokomocutie88r1
Note $\mathrm{ln}2=a,\mathrm{ln}3=b,\mathrm{ln}x=u,\mathrm{ln}y=v.$
Logarithm equations are obtained:
${a}^{2}+au={b}^{2}+bv$
and
$bu=av.$
With
$v=\frac{b}{a}u$
find
$u=-a.$
Conclusion:
${x}_{0}=\frac{1}{2}.$