Question

How do you solve for x?: `e^(-2x)+e^x = 1/3?`

Answer 1

I found: `-ln[3(e^-2+1)]=-ln(3)-ln(e^-2+1)`

Explanation:

Collect `e^x` on the left:
`e^x(e^-2+1)=1/3` rearrange:
`e^x=1/(3(e^-2+1))`
apply the `ln` on both sides:
`ln[e^x]=ln[1/(3(e^-2+1))]`
so:
`x=ln[1/(3(e^-2+1))]`
use the properties of log relating sums and subtractions to multiplications and divisions to get:
`x=ln(1)-ln[3(e^-2+1)]=0-ln(3)-ln(e^-2+1)`