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

1 Answer
Feb 7, 2017

Answer:

#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#