How do you solve $3 ln x + ln 5 = 7$ $3lnx+ln5=7$ ?

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Answer 1 · 2015-11-26T18:44:22

Use the properties of logs ...

$3 ln x + ln 5 = ln [(5) (x^{3})]$ $3lnx+ln5=ln[(5)(x^3)]$

Now, exponentiate both sides of the equation ...

$e^{ln [(5) (x^{3})]} = e^{7}$ $e^ln[(5)(x^3)]=e^7$

Simplify and solve for x...

$5 x^{3} = e^{7}$ $5x^3=e^7$

$x^{3} = \frac{e^{7}}{5}$ $x^3=(e^7)/5$

${(x^{3})}^{\frac{1}{3}} = x = {[\frac{e^{7}}{5}]}^{\frac{1}{3}} \approx 6.031$ $(x^3)^(1/3)=x=[(e^7)/5]^(1/3)~~6.031$

hope that helped

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How do you solve 3lnx+ln5=73lnx+ln5=73lnx+ln5=7?