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

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Answer 1 · 2015-12-21T04:49:51

$x = e^{7}$ $x=e^7$

Divide both sides by $5$ $5$ .

$ln x = 7$ $lnx=7$

Undo the natural logarithm by exponentiating both sides.

$e^{ln x} = e^{7}$ $e^(lnx)=e^7$

$x = e^{7}$ $x=e^7$

This could also be solved by rewriting $5 ln x$ $5lnx$ using logarithm rules.

$ln (x^{5}) = 35$ $ln(x^5)=35$

$e^{ln (x^{5})} = e^{35}$ $e^(ln(x^5))=e^35$

$x^{5} = e^{35}$ $x^5=e^35$

${(x^{5})}^{\frac{1}{5}} = {(e^{35})}^{\frac{1}{5}}$ $(x^5)^(1/5)=(e^35)^(1/5)$

$x = e^{7}$ $x=e^7$

How do you solve 5lnx=355ln x = 35?