Question

How do you solve `e^x(2e^x-1) = 10`?

Answer 1

Real solution: `x=ln(5/2)`

Complex solutions:

`x=ln(5/2)+2npii`

`x=ln(2)+(2n+1)pii`

Explanation:

Given:

`e^x(2e^x-1) = 10`

We can treat this as a quadratic in `e^x` and factor it using an AC method:

Transposing and subtracting `10` from both sides, this becomes:

`0 = e^x(2e^x-1)-10`

`color(white)(0) = 2(e^x)^2-(e^x)-10`

`color(white)(0) = (2(e^x)^2-5(e^x))+(4(e^x)-10)`

`color(white)(0) = e^x(2e^x-5)+2(2e^x-5)`

`color(white)(0) = (e^x+2)(2e^x-5)`

So:

`e^x = -2" "` or `" "e^x = 5/2`

If `x` is Real then `e^x > 0`. So the only Real solution to the given equation is given by taking the natural logarithm of both sides of the second equation to find:

`x = ln (5/2)`

Note however that `e^(ipi) = -1`, and hence there are complex solutions:

`x = ln(5/2) + 2npi i`

`x = ln(2) + (2n+1)pi i`

for any integer `n`