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.

Related questions
See all questions in Logarithmic Models
Impact of this question
2513 views around the world
You can reuse this answer
Creative Commons License