Question

How do you solve `e^x>1.6`?

Answer 1

`x > ln1.6`, with `ln1.6 ~~ 0.470003629`.

Explanation:

Since `lnx` is a constantly increasing function, for any `a > b` it holds that `lna > lnb`. Applying this to the relationship:

`ln e^x > ln1.6`, and we know that `ln e^x = x`

If even further confirmation is required, from logarithm properties:

`ln e^x = xlne = x* 1 = x`