find the exact value of x for which

ln(x^2+4x+3)-ln(x^2+x)=4

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- Jun 5th 2008, 11:27 AMoutlawneed help ASAP
find the exact value of x for which

ln(x^2+4x+3)-ln(x^2+x)=4 - Jun 5th 2008, 11:33 AMo_O
$\displaystyle \ln (a) - \ln (b) = \ln \left(\frac{a}{b}\right)$

$\displaystyle \ln (x^2+4x+3) - \ln (x^2+x) = 4$

$\displaystyle \ln \left(\frac{x^{2} + 4x + 3}{x^{2} + x}\right) = 4$

Simplify the inside by factoring and canceling the common factor.

This should help you right after: $\displaystyle \ln (a) = b \iff \log_{e}a = b \iff a^{b} = e$ where e is the constant 2.718...

Then it should be a matter of simple algebra to solve for x. - Jun 5th 2008, 11:35 AMIsomorphism