Question

How do you solve `e^(x-2) = 5`?

Answer 1

`x=2+ln5 = 3.6094`

Explanation:

Take the natural logarithm on both sides and use the laws of logs to obtain

`ln(e^(x-2))=ln5`

`therefore (x-2)lne=ln5`

`therefore x=2+ln5`

`=3.6094`