How do you solve #e^[2 ln x – ln (x^2 + x - 3)] = 1#? Precalculus Properties of Logarithmic Functions Functions with Base b 1 Answer A. S. Adikesavan · David B. May 6, 2016 #x = 3# Explanation: Use #m ln n=ln m^n and e^log a=a# The given equation is #e^(ln x^2-ln(x^2+x-3))=1# #e^(ln x^2)/e^ln(x^2+x-3)=1# #x^2/(x^2+x-3)=1# #x^2=x^2+x-3# #x=3# Answer link Related questions What is the exponential form of #log_b 35=3#? What is the product rule of logarithms? What is the quotient rule of logarithms? What is the exponent rule of logarithms? What is #log_b 1#? What are some identity rules for logarithms? What is #log_b b^x#? What is the reciprocal of #log_b a#? What does a logarithmic function look like? How do I graph logarithmic functions on a TI-84? See all questions in Functions with Base b Impact of this question 3651 views around the world You can reuse this answer Creative Commons License