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

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

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

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Answer 1 · 2018-05-01T13:37:19

$x = \frac{1}{4} ln (3 \pm \sqrt{6})$ $x=1/4ln(3+-sqrt(6))$

$x_{1} \approx 0.42$ $x_1~~0.42$

$x_{2} \approx - 0.15$ $x_2~~-0.15$

Explanation:

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

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

Notice how,

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

$\Leftrightarrow e^{4 x} + 3 e^{(4 x) \cdot - 1} = 6$ $<=>e^(4x)+3e^((4x)*-1)=6$

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

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

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

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

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

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

Using the quadratic formula, we get:

$u = \frac{6 \pm \sqrt{36 - 4 \cdot 1 \cdot 3}}{2}$ $u=(6+-sqrt(36-4*1*3))/2$

$= \frac{6 \pm \sqrt{36 - 12}}{2}$ $=(6+-sqrt(36-12))/2$

$= \frac{6 \pm \sqrt{24}}{2}$ $=(6+-sqrt(24))/2$

$= \frac{6 \pm 2 \sqrt{6}}{2}$ $=(6+-2sqrt(6))/2$

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

$e^{4 x} = \frac{6 \pm 2 \sqrt{6}}{2}$ $e^(4x)=(6+-2sqrt(6))/2$

$= 3 \pm \sqrt{6}$ $=3+-sqrt(6)$

Take natural logs on both sides.

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

$4 x ln e = ln (3 \pm \sqrt{6})$ $4xlne=ln(3+-sqrt(6))$

$4 x = ln (3 \pm \sqrt{6})$ $4x=ln(3+-sqrt(6))$

$:.x=1/4ln(3+-sqrt(6))$

Answer 2 · 2018-05-01T14:06:35

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

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 $e^(4x)$ is common to each term in the expression, so you can write:
$y=e^(4x)$

That gives $y -6+3y^(-1)=0$
Get rid of y in the numerator place by multiplying each term with y. This gives the following 2nd degree equation:
$y^2 -6y+3=0$
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 $y=e^(4x)$ , you find that
$4x=ln(y)$ , i.e.
$x_1 =ln(y_1)/4$ = ln(3+√6) (=0.42)
$x_2 =ln(y_2)/4$ = ln(3-√6) (=-0.15)

Check:
$e^(4x_1) + 3e^(-4x_1) - 6$
= $e^(1.70) + 3e^(-1.70) - 6 = 0$ -> check
$e^(4x_2) + 3e^(-4x_2) - 6$
= $e^(-0.60) + 3e^(0.60) - 6 = 0$ -> check

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

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

2 Answers

Explanation:

Explanation:

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