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

1 Answer
iceman
Sep 2, 2015

#x≈14.37#

Explanation:

#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.

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