# How to solve for t? 3e^t=5+8e^(-t)

Apr 25, 2018

Solution: $t \approx 0.98$

#### Explanation:

$3 {e}^{t} = 5 + 8 {e}^{- t}$ Multiplying by ${e}^{t}$ on both sides we get,

$3 {e}^{2 t} = 5 {e}^{t} + 8$ or

$3 {e}^{2 t} - 5 {e}^{t} - 8 = 0$ or

$3 {e}^{2 t} + 3 {e}^{t} - 8 {e}^{t} - 8 = 0$ or

$3 {e}^{t} \left({e}^{t} + 1\right) - 8 \left({e}^{t} + 1\right) = 0$ or

$\left({e}^{t} + 1\right) \left(3 {e}^{t} - 8\right) = 0$

:. e^t = -1 or 3 e^t =8 ; e^t cannot be negative number

$\therefore {e}^{t} = \frac{8}{3}$, taking natural log on both sides we get,

$t \ln e = \ln 8 - \ln 3 \therefore t = \ln 8 - \ln 3 \approx 0.98 \left(2 \mathrm{dp}\right) \left[\ln e = 1\right]$

Solution: $t \approx 0.98$ [Ans]

Apr 25, 2018

$t = \ln \left(\frac{8}{3}\right) \approx 0.9808$

#### Explanation:

Here,

$3 {e}^{t} = 5 + 8 {e}^{-} t$

$\implies 3 {e}^{t} = 5 + \frac{8}{{e}^{t}}$

$\implies 3 {\left({e}^{t}\right)}^{2} = 5 {e}^{t} + 8$

$\implies 3 {\left({e}^{t}\right)}^{2} - 5 {e}^{t} - 8 = 0$

$\implies 3 {\left({e}^{t}\right)}^{2} + 3 {e}^{t} - 8 {e}^{t} - 8 = 0$

$\implies 3 {e}^{t} \left({e}^{t} + 1\right) - 8 \left({e}^{t} + 1\right) = 0$

$\implies \left({e}^{t} + 1\right) \left(3 {e}^{t} - 8\right) = 0$

$\implies {e}^{t} + 1 = 0 \mathmr{and} 3 {e}^{t} - 8 = 0$

$\implies {e}^{t} = - 1 \mathmr{and} {e}^{t} = \frac{8}{3}$

$\left(i\right) {e}^{t} = - 1 < 0 \implies {e}^{t} \notin {R}^{+}$

$\left(i i\right) {e}^{t} = \frac{8}{3} \implies t = \ln \left(\frac{8}{3}\right) \approx 0.9808$