How do you solve #e^x>30#?

1 Answer
Jul 8, 2018

Answer:

#x > ln(30)#

Explanation:

It may be helpful to look at a graph of #e^x#. Notice that because it is an exponential function with a base greater than 1, it is monotonically increasing, so:

#e^a > e^b => a > b#

To clarify, I'll use a concrete example.

#e^2 > e^0 => 2 > 0#

Getting back to our original problem, we want to find when #e^x# is greater than #30#. Rewrite #30# as #e^ln(30)#

#e^x > e^ln(30)#

But if #e^x > e^ln(30)#:

#x > ln(30)#

Technically, you could also take the natural logarithm on both sides since that is also monotonically increasing.