How do you solve #log_7 7(x+1) + log_7 (x-5)=1#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Vinícius Ferraz Nov 1, 2015 #x = 2 ± 2sqrt 2# Explanation: #log a + log b = log (ab)#, then #log_7 [7(x+1)(x-5)] = 1# #log_b a = c \Rightarrow b^c = a#, then #7^1 = 7(x+1)(x-5)# #1 = (x+1)(x-5)#, continue with Baskara 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 4864 views around the world You can reuse this answer Creative Commons License