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

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Answer 1 · 2015-12-31T00:50:31

$x = \frac{e^{2}}{3}$ $x=e^2/3$

Method One

Divide both sides by $3$ $3$ .

$ln 3 x = 2$ $ln3x=2$

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

$e^{ln 3 x} = e^{2}$ $e^(ln3x)=e^2$

$3 x = e^{2}$ $3x=e^2$

$x = \frac{e^{2}}{3}$ $x=(e^2)/3$

Method Two

Rewrite the original expression using logarithm rules.

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

$ln (27 x^{3}) = 6$ $ln(27x^3)=6$

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

$27 x^{3} = e^{6}$ $27x^3=e^6$

$x^{3} = \frac{e^{6}}{27}$ $x^3=(e^6)/27$

${(x^{3})}^{\frac{1}{3}} = {⎛ ⎝ \frac{{(e^{2})}^{3}}{3^{3}} ⎞ ⎠}^{\frac{1}{3}}$ $(x^3)^(1/3)=(((e^2)^3)/(3^3))^(1/3)$

$x = \frac{e^{2}}{3}$ $x=(e^2)/3$

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