How do you solve #2^(x-3) = 5 ^ (x+7)#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Tony B May 4, 2016 #=>x~~-14.5647" to 4 decimal places")# Explanation: Taking logs #(x-3)ln(2)=(x+7)ln(5)# #=>x-3=(x+7)ln(5)/ln(2)# Let #ln(5)/ln(2)=A# #x-3=(x+7)A# #=>x-Ax=7A+3 larr# Corrected from -3 Tony B #=>x=(7A+3)/(1-A)# #cancel(=>x~~-10.0259" to 4 decimal places")# #=>x~~-14.5647" to 4 decimal places")# 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 1241 views around the world You can reuse this answer Creative Commons License