Logs rules and Solving I've got the equation :
$$-1=\frac{-8e^{-t} + 3e^t}{2e^t}$$
I've moved some stuff around to get :
$5e^t = 8e^{-t} $
But not sure where to go from here.
Thanks for any help
 A: cross multiply you get $-2e^t = -8e^{-t} + 3e^{t}$ and then you get $5e^t = 8e^{-t}$ like you said then you take natural $\log$ both sides to get $\ln(5e^t) = \ln(8e^{-t})$ and notice that $\ln(5e^t) = \ln(5) + \ln(e^t) =\ln(5) + t$ and also notice that $\ln(8e^{-t}) = \ln(8) + \ln(e^{-t}) = \ln(8) -t$ and so we get $$\ln(5) + t = \ln(8) -t$$ and so $$ 2t = \ln(8) - \ln(5)$$ and so $$2t = \ln(\frac{8}{5})$$ and so $$t = 0.5 \ln(\frac{8}{5})$$
Notice that $\ln(8) = \ln(2^3) = 3\ln(2)$ as well
Notice that I used these these properties in my answer
(1) $\large{\color{red}{\ln(ab) = \ln(a) + \ln(b)}}$
(2) $\large{\color{red}{ln(\frac{a}{b}) = \ln(a) - \ln(b)}}$
(3) $\large{\color{red}{\ln(e^m) = m}}$
(4) $\large{\color{red}{\ln(a^b) = b\ln(a)}}$
A: $5e^t=8e^{-t}$
Take the natural logarithm of both sides and use idintities $\log(ab)=\log(a)+\log(b)$ and $\log(a^b)=b\log(a)$
$t+\ln(5)=3\ln(2)-t$
$2t=3\ln(2)-\ln(5)$
$t=\frac{3\ln(2)}{2}-\frac{\ln(5)}{2}$
A: Or dividing the right side since there is only a single term in the denominator: 
$ -1 = -4e^{-2t} + 3/2 $
$ 4e^{-2t} = 5/2 $
$ e^{-2t} = 5/8  $    
Now, after taking the natural log of both sides:
$ -2t = ln (5/8) $
$ t = -\frac{ln(5/8)}{2} $
This is equivelent to the two previous answers.
