Question

How do I solve for `x`? `2e^{2x}-5e^x-3=0` This problem is about the natural logarithm function.

Answer 1

`x = log_e3`

Explanation:

Making `e^x = y` and substituting we have the equivalent problem.

`2 y^2-5y-3=0`

Solving for `y` gives

`y = {-1/2,3}` we take the consistent one which is

`y = 3 = e^x` because `e^x > 0 forall x in RR` and solving for `x` gives

`x = log_e3`