Question

How do you solve for x in `2e^(x-2)=e^x + 7`?

Answer 1

`x=ln7-ln(2e^(-2)-1)`

Explanation:

`2e^(x-2)=e^x+7`

or `2e^x/e^2=e^x+7`

or `2e^x/e^2-e^x=7`

or `e^x(2e^(-2)-1)=7`

or `e^x=7/((2e^(-2)-1)`

or `x=ln(7/((2e^(-2)-1)))`

or `x=ln7-ln(2e^(-2)-1)`