How do you solve for x in #3ln3x=6#?

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Answer 1 · 2015-12-31T00:50:31

#x=e^2/3#

Method One

Divide both sides by #3#.

#ln3x=2#

To undo the natural logarithm, exponentiate both sides with base #e#.

#e^(ln3x)=e^2#

#3x=e^2#

#x=(e^2)/3#

Method Two

Rewrite the original expression using logarithm rules.

#ln((3x)^3)=6#

#ln(27x^3)=6#

#e^(ln(27x^3))=e^6#

#27x^3=e^6#

#x^3=(e^6)/27#

#(x^3)^(1/3)=(((e^2)^3)/(3^3))^(1/3)#

#x=(e^2)/3#