How do you solve #(ln(x))^2 + ln(x) - 6 = 0#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer maganbhai P. Jun 5, 2018 #x=e^(-3)=1/(e^3) or x=e^2# Explanation: We know that, #(1)log_a y=X <=>y=a^X, AAa in RR^+ -{1},yinRR^+,X inRR# #(2)color(red)(ln(y)=X<=>y=e^X ),where,yinRR^+ andcolor(red)( X inRR# Here, #(ln(x))^2+ln(x)-6=0# Let, #ln(x)=m# So, #m^2+m-6=0# #m^2+3m-2m-6=0# #m(m+3)-2(m+3)=0# #(m+3)(m-2)=0# #m+3=0 or m-2=0# #m=-3 or m=2# Subst. back, #m=ln(x)# #:.lnx=-3 or lnx=2,where,color(red)( (-3) and 2inRR# #=>x=e^(-3) or x=e^2# 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 4833 views around the world You can reuse this answer Creative Commons License