How do you solve #5ln x = 35#?

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Answer 1 · 2015-12-21T04:49:51

#x=e^7#

Divide both sides by #5#.

#lnx=7#

Undo the natural logarithm by exponentiating both sides.

#e^(lnx)=e^7#

#x=e^7#

This could also be solved by rewriting #5lnx# using logarithm rules.

#ln(x^5)=35#

#e^(ln(x^5))=e^35#

#x^5=e^35#

#(x^5)^(1/5)=(e^35)^(1/5)#

#x=e^7#