How do you solve #log _2 (x+2) - log _2 (x-5) = 3#? Precalculus Properties of Logarithmic Functions Functions with Base b 1 Answer A. S. Adikesavan May 20, 2016 6 Explanation: Use, #log_bm-log_bn=log_b(m/n)# and, if #log_b a=c# then #a=b^c# #log_2 (x+2)-log_2 (x-5)# #=log_2((x+2)/(x-5))=3#. Inversely, (x+2)/(x-5)= 2^3=8#. Solving,. x=6.. 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 1368 views around the world You can reuse this answer Creative Commons License