How do you evaluate #e^(In8)#?

1 Answer
Jul 5, 2016

Answer:

#e^ln(8) = 8#

Explanation:

Assuming the intended question is to evaluate #e^ln(8)#:

The base #b# logarithm is the inverse of an exponential function with base #b#. Specifically, it is the value to which #b# must be raised to obtain the argument of the function.

#log_b(x) = y iff x = b^y#

Note, then, that by definition: #a^(log_a(x)) = a^y = x#

(Note that it should make intuitive sense as to why #a^(log_a(x))=x#, as #log_a(x)# is the power to which we would need to raise #a# to obtain #x#, and we are raising #a# to that power)

The natural logarithm, or #ln#, is the logarithm with the base #e#. With that, we have

#e^ln(8) = e^(log_e(8)) = 8#