Question

How do you find the primitive of `5^-x` given that you have to use the identity a=`e^(lna)` to express it as a power of e?

Answer 1

`int 5^(-x)dx = -5^(-x)/ln5 +C`

Explanation:

Note that:

`5 = e^(ln5)`

so that:

`5^(-x) = (e^(ln5))^(-x) = e^(-xln5)`

Thus:

`int 5^(-x)dx = int e^(-xln5) dx = -e^(-xln5)/ln5 + C = -5^(-x)/ln5 +C`