My attempt went as follows:

ln e ^ (2x) + ln 2 e ^ x = ln 8

No, that is not correct.

ln(a +b) is not ln(a) +ln(b).

Rather, ln(a*b) = ln(a) +ln(b)

By the way, your equation do not need logs yet.

If you look closely, it is a quadratic equation in e^x

e^(2x) +2e^x = 8

Or, that is,

(e^x)^2 +2e^x -8 = 0

It is like having

y^2 +2y -8 = 0

if y = e^x

So, factoring that,

(y +4)(y-2) = 0

y +4 = 0

y = -4

Or, back to the e^x for y,

e^x = -4

Take the ln of both sides,

ln(e^x) = ln(-4)

x*ln(e) = ln(-4)

x = ln(-4)

But there are no logs of negative numbers (no real number logarithmns for negative numbers), so reject e^x = -4

y -2 = 0

Or,

e^x -2 = 0

e^x = 2

x = ln(2) = 0.693147181 -----------------answer.