E^4x + 3e^-4x = 6 x=?

e^4x + 3e^-4x = 6

May 1, 2018

$x = \frac{1}{4} \ln \left(3 \pm \sqrt{6}\right)$

${x}_{1} \approx 0.42$

${x}_{2} \approx - 0.15$

Explanation:

Given: ${e}^{4 x} + 3 {e}^{- 4 x} = 6$.

Let $u = {e}^{4 x}$.

Notice how,

${e}^{4 x} + 3 {e}^{- 4 x} = 6$

$\iff {e}^{4 x} + 3 {e}^{\left(4 x\right) \cdot - 1} = 6$

Therefore, replacing $4 x$ by $u$, we get:

$u + 3 {u}^{-} 1 = 6$

$u + \frac{3}{u} = 6$

Multiply by ${u}^{2}$ to get:

${u}^{2} + 3 = 6 u$

${u}^{2} - 6 u + 3 = 0$

Using the quadratic formula, we get:

$u = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 3}}{2}$

$= \frac{6 \pm \sqrt{36 - 12}}{2}$

$= \frac{6 \pm \sqrt{24}}{2}$

$= \frac{6 \pm 2 \sqrt{6}}{2}$

Replacing $u = {e}^{4 x}$ back, we get:

${e}^{4 x} = \frac{6 \pm 2 \sqrt{6}}{2}$

$= 3 \pm \sqrt{6}$

Take natural logs on both sides.

$\ln \left({e}^{4 x}\right) = \ln \left(3 \pm \sqrt{6}\right)$

$4 x \ln e = \ln \left(3 \pm \sqrt{6}\right)$

$4 x = \ln \left(3 \pm \sqrt{6}\right)$

$\therefore x = \frac{1}{4} \ln \left(3 \pm \sqrt{6}\right)$

May 1, 2018

Do you mean the following expression? ${e}^{4 x} + 3 {e}^{- 4 x} = 6$

Explanation:

It is important that the expression becomes right. But rather than do the whole calculus, I'll lead you on your way.

To solve such an expression, please note that ${e}^{4 x}$ is common to each term in the expression, so you can write:
$y = {e}^{4 x}$

That gives $y - 6 + 3 {y}^{- 1} = 0$
Get rid of y in the numerator place by multiplying each term with y. This gives the following 2nd degree equation:
${y}^{2} - 6 y + 3 = 0$
You solve this equation the normal way, which gives you two solutions
y_1=3+√6 (= 5.45)
y_2=3 -√6 (=0.55)

As $y = {e}^{4 x}$, you find that
$4 x = \ln \left(y\right)$, i.e.
${x}_{1} = \ln \frac{{y}_{1}}{4}$ = ln(3+√6) (=0.42)
${x}_{2} = \ln \frac{{y}_{2}}{4}$ = ln(3-√6) (=-0.15)

Check:
${e}^{4 {x}_{1}} + 3 {e}^{- 4 {x}_{1}} - 6$
=${e}^{1.70} + 3 {e}^{- 1.70} - 6 = 0$ -> check
${e}^{4 {x}_{2}} + 3 {e}^{- 4 {x}_{2}} - 6$
=${e}^{- 0.60} + 3 {e}^{0.60} - 6 = 0$ -> check