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April 20, 2014

April 20, 2014

Posted by **Erik** on Wednesday, April 27, 2011 at 3:04pm.

Solve for x

lnx-lnx^2+ln5 = 0

Simplify the answer

- Math -
**MathMate**, Wednesday, April 27, 2011 at 3:16pmUse the law of logarithms:

ln(a)+ln(b)=ln(ab)

ln(a)-ln(b)=ln(a/b)

kln(a)= ln(a^{k})

ln(1)=0

ln(1/x)=-ln(x)

so

ln(x)-ln(x²)+ln(5)=0

ln(x/x²)+ln(5)=0

ln(1/x)+ln(5)=0

-ln(x)+ln(5)=0

ln(x)=ln(5)

e^(ln(x))=e^(ln(5))

x=5

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