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.

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

$ln(27x^3)=6$

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

$27x^3=e^6$

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

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

$x=(e^2)/3$

How do you solve for x in 3ln3x=63ln3x=6?