How do you solve #e^ { \ln x } = 21#?

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Answer 1 · 2018-04-17T05:19:10

x=21, e #e^(lnx) = x#

x=21, e #e^(lnx) = x# by the rules of logarithms and exponential funcs

Answer 2 · 2018-04-17T05:21:48

#x=21#

Given: #e^(lnx)=21#

Notice that #e^(lnx)=x#.

So, we get: #x=21#