How do you solve #5log_10(x-2)=11#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Somebody N. Jul 3, 2018 #color(blue)(x=10^(11/5)+2~~160.4893192)# Explanation: #5log_(10)(x-2)=11# Divide by 5: #log_(10)(x-2)=11/5# Raising the base to these: #10^(log_(10)(x-2))=10^(11/5)# #x-2=10^(11/5# #x=10^(11/5)+2~~160.4893192# 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 2476 views around the world You can reuse this answer Creative Commons License