How do you solve $lnx + ln(x^2 + 1) = 8$ ?

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Answer 1 · 2015-09-02T12:20:16

$x≈14.37$

$ln(x) + ln(x^2 + 1) = 8=>$ use: $log(a) + log(b) = log(ab)$

$ln[x(x^2 + 1)]=8=>$ expand inside the brackets:

$ln(x^3 + x)=8=>$ if: $ln(x)=y <=> x= e^y:$

$x^3+x=e^8=>$ using an equation solver:

$x≈14.37$

The other two roots are not real, they are complex.

How do you solve lnx + ln(x^2 + 1) = 8lnx + ln(x^2 + 1) = 8?