How do you solve $ln 4 x + ln x = 9$ $ln4x+lnx=9$ ?

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Answer 1 · 2016-11-16T18:15:52

Use the product rule for logarithms you get

$ln 4 x + ln x = 9$ $ln4x+lnx=9$

$ln 4 x^{2} = 9$ $ln4x^2=9$

$4 x^{2} = e^{9}$ $4x^2=e^9$

${(2 x)}^{2} = e^{9}$ $(2x)^2=e^9$

$2 x = \sqrt{e^{9}}$ $2x=sqrt(e^9)$

$x = \frac{\sqrt{e^{9}}}{2}$ $x=sqrt(e^9)/2$

Remark : from the defination of the logarithm we have $x > 0$ $x>0$

How do you solve ln4x+lnx=9ln4x+lnx=9?