How do you solve #7^x = 80#?

1 Answer
Mar 22, 2016

well, by inspection we know that #7^2=49 and 7^3=343#
so this means that the exponent 'x' must be between 2 and 3 (and closer to 2 than to 3).

so we convert from exponent form to log form and we obtain: #log_7(80)=x# which can be solved on a calculator or by using the change of base rule: #log80/log7#
or approximately 2.25