Question

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

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

Answer 1

`x=1/4ln(3+-sqrt(6))`

`x_1~~0.42`

`x_2~~-0.15`

Explanation:

Given: `e^(4x)+3e^(-4x)=6`.

Let `u=e^(4x)`.

Notice how,

`e^(4x)+3e^(-4x)=6`

`<=>e^(4x)+3e^((4x)*-1)=6`

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

`u+3u^-1=6`

`u+3/u=6`

Multiply by `u^2` to get:

`u^2+3=6u`

`u^2-6u+3=0`

Using the quadratic formula, we get:

`u=(6+-sqrt(36-4*1*3))/2`

`=(6+-sqrt(36-12))/2`

`=(6+-sqrt(24))/2`

`=(6+-2sqrt(6))/2`

Replacing `u=e^(4x)` back, we get:

`e^(4x)=(6+-2sqrt(6))/2`

`=3+-sqrt(6)`

Take natural logs on both sides.

`ln(e^(4x))=ln(3+-sqrt(6))`

`4xlne=ln(3+-sqrt(6))`

`4x=ln(3+-sqrt(6))`

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

Answer 2

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