# How do you solve e^x+4=1/(e^(2x)) ?

Nov 28, 2015

Express as a cubic in ${e}^{x}$, solve that, then take natural log.

#### Explanation:

Multiply both sides by ${e}^{2 x}$ to get:

${e}^{3 x} + 4 {e}^{2 x} = 1$

Subtract $1$ from both sides to get:

${e}^{3 x} + 4 {e}^{2 x} - 1 = 0$

Let $t = {e}^{x}$.

${t}^{3} + 4 {t}^{2} - 1 = 0$

This cubic has three Real roots, which are all irrational, but only one is positive.

graph{x^3+4x^2-1 [-10.58, 9.42, -1.36, 8.64]}

Solve the cubic by your favourite method to find:

${t}_{1} \approx 0.472833909$

Then $x = \ln \left({t}_{1}\right) \approx - 0.749011$

Nov 28, 2015

Let $t = {e}^{-} x$ to get a cubic ${t}^{3} - 4 t - 1 = 0$ with $3$ Real roots, one of which is positive, giving $t \approx 2.11490754$ and

$x = - \ln \left(t\right) \approx - 0.749011$

#### Explanation:

Divide both sides of the equation by ${e}^{x}$ to get:

$1 + \frac{4}{e} ^ x = \frac{1}{{e}^{x}} ^ 3$

Subtract the left hand side from the right to get:

${\left(\frac{1}{e} ^ x\right)}^{3} - 4 \left(\frac{1}{e} ^ x\right) - 1 = 0$

Let $t = \frac{1}{e} ^ x$ to get:

${t}^{3} - 4 t - 1 = 0$

This cubic has three Real roots, one of which is positive.

graph{x^3-4x-1 [-10, 10, -5, 5]}

Find by your favourite method (*):

${t}_{1} \approx 2.11490754$

Then $x = - \ln \left({t}_{1}\right) \approx - 0.749011$

(*) For example, for a cubic with $3$ Real roots that is already in the form ${t}^{3} + p t + q = 0$ you can use the trigonometric formula:

${t}_{k} = 2 \sqrt{- \frac{p}{3}} \cos \left(\frac{1}{3} \arccos \left(\frac{3 q}{2 p} \sqrt{- \frac{3}{p}}\right) - \frac{2 \pi k}{3}\right)$

with $k = 0 , 1 , 2$

choosing $k$ to give you the positive root.