How do you solve #lnx + ln(x^2 + 1) = 8#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 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. Answer link Related questions What is a logarithmic model? How do I use a logarithmic model to solve applications? What is the advantage of a logarithmic model? How does the Richter scale measure magnitude? What is the range of the Richter scale? How do you solve #9^(x-4)=81#? How do you solve #logx+log(x+15)=2#? How do you solve the equation #2 log4(x + 7)-log4(16) = 2#? How do you solve #2 log x^4 = 16#? How do you solve #2+log_3(2x+5)-log_3x=4#? See all questions in Logarithmic Models Impact of this question 2805 views around the world You can reuse this answer Creative Commons License