Need to solve the following for x
e^(3x-4) = e^(1-2x) Do i figure out the logs first or do i combine the x terms?
ln(x) = 2ln(2)+ln(15)-ln(10) (for x>0)
ln(3x-2) = 3 (x>2/3)
thanks for your help
Hi tim_mannire,
Just take the natural log of both sides.
$\displaystyle e^{3x-4}=e^{1-2x}$
$\displaystyle \ln e^{3x-4}=\ln e^{1-2x}$
$\displaystyle 3x-4=1-2x$
$\displaystyle x=e^{2\ln2+\ln15-\ln10}$
$\displaystyle e^3=3x-2$