Question

2e^(2x)-7e^(x)=15

Answer 1

this is an equation that is reducible to a quadratic

\[2(e^x)^2 - 7e^x - 15 = 0\]

so it can be solved by factoring

\[(2e^x + 3)(e^x - 5) = 0\]

so to find the value of x that make the equation true, you'll need to solve

\[2e^x + 3 = 0\]

this has no solution as you can't take the log of a negative number

so all you need to do is solve

\[e^x - 5 = 0\]