How do you solve log_2 (x+1) - log_2 (x-1) = 4log2(x+1)−log2(x−1)=4? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alan P. Dec 27, 2015 x=17/15x=1715 Explanation: log_2(x+1)-log_2(x-1) = log_2((x+1)/(x-1))log2(x+1)−log2(x−1)=log2(x+1x−1) 2^4 = 16 rArr 4=log_2(16)24=16⇒4=log2(16) Therefore color(white)("XXX")log_2(x+1)-log_2(x-1)=4XXXlog2(x+1)−log2(x−1)=4 is equivalent to color(white)("XXX")log_2((x+1)/(x-1))=log_2(16)XXXlog2(x+1x−1)=log2(16) color(white)("XXX")(x+1)/(x-1) = 16XXXx+1x−1=16 color(white)("XXX")x+1 = 16x-16XXXx+1=16x−16 color(white)("XXX")-15x = -17XXX−15x=−17 color(white)("XXX")x=17/15XXXx=1715 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)=819x−4=81? How do you solve logx+log(x+15)=2logx+log(x+15)=2? How do you solve the equation 2 log4(x + 7)-log4(16) = 22log4(x+7)−log4(16)=2? How do you solve 2 log x^4 = 162logx4=16? How do you solve 2+log_3(2x+5)-log_3x=42+log3(2x+5)−log3x=4? See all questions in Logarithmic Models Impact of this question 1790 views around the world You can reuse this answer Creative Commons License