Where does the derivative of e^(2x)+e^(-x) equal zero?

Question

I found the derivative to be 2e^(2x) - e^(-x), but I can't figure out how to solve for x.

Answer 1

The derivative is, as you note, 2e^(2x) - e^(-x). So set this equal to zero:

2e^(2x) - e^(-x) = 0

Multiply both sides by e^x:

2e^(3x) - 1 = 0

Add 1 to both sides, then divide by 2:

e^(3x) = 1/2

Take the natural log of both sides:

3x = ln (1/2)

Divide by 3:

x = 1/3 ln (1/2)

And at this point we are done. However, we can simplify the answer further:

x = -1/3 ln 2
x = - ln (∛2)

Check of result: plugging in -ln (∛2) yields:

2e^(-2 ln (∛2)) - e^(ln (∛2))
2/((∛2)²) - ∛2
∛2 - ∛2
0

So our result is correct.

Answer 2

y = e^(2x)+e^(-x)
y' = 2 e^(2x) - e^(-x) =0

2 e^(2x) = e^(-x)

Take the log of both sides

Ln (2 e^(2x)) = Ln(e^(-x))
Ln(2) + 2x = -x
3x = Ln2
x = (Ln2)/3

Answer 3

try this

2e^(2x) - e^(-x) = 0

2e^(2x) = e^(-x)

Ln(2e^2x) = Ln(e^(-x))

Ln2 + 2x = -x

3x = Ln2

x = (Ln2)/3

Check this