# need help ASAP

• Jun 5th 2008, 11:27 AM
outlaw
need 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 AM
o_O
$\ln (a) - \ln (b) = \ln \left(\frac{a}{b}\right)$

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

Simplify the inside by factoring and canceling the common factor.

This should help you right after: $\ln (a) = b \iff \log_{e}a = b \iff a^{b} = e$ 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 AM
Isomorphism
Quote:

Originally Posted by outlaw
find the exact value of x for which

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

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

ln((x+3)(x+1))-ln(x(x+1))=4

ln(x+3) + ln(x+1)-(ln x + ln(x+1))=4

ln(x+3) -ln x = 4

ln((x+3)/x) = 4

(x+3)/x = e^4

x = 3/(e^4 - 1)