Logs rules and Solving

Question

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

Answer 1

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) $\large{\color{red}{\ln(a^b) = b\ln(a)}}$

Answer 2

$5e^t=8e^{-t}$

Take the natural logarithm of both sides and use idintities $\log(ab)=\log(a)+\log(b)$ and $\log(a^b)=b\log(a)$

$t+\ln(5)=3\ln(2)-t$

$2t=3\ln(2)-\ln(5)$

$t=\frac{3\ln(2)}{2}-\frac{\ln(5)}{2}$

Answer 3

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.