How do you solve #2 ln (7x) = 4#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer maganbhai P. Aug 9, 2018 #:.x=e^2/7# Explanation: Here , #2ln(7x)=4# #where , x in RR^+# #:.ln(7x)^2=4...to[becauselna^n=nlna]# #:.(7x)^2=e^4....to[becauselny=k<=>y=e^k ]# #:.7x=e^2# #:.x=e^2/7# Check : #LHS=2ln(7x)=2ln(7*e^2/7)=2lne^2=4lne=4(1)=4# 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 3050 views around the world You can reuse this answer Creative Commons License