How do you solve #ln(e^(7x)) = 15 #?

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Answer 1 · 2018-03-02T18:44:01

#x=15/7#

This is for natural logarithms.

#ln(e^(7x))=15#

#ln(a^b)=bln(a)#

Then:

#ln(e^(7x))=15#

#7xln(e)=15#

#ln(e)=1# ( The logarithm of the base is always 1 )

#7x=15=>x=15/7#