How do you solve #6^x+10=47#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Monzur R. Feb 19, 2017 #x=log_6 37# Explanation: #6^x+10=47# #6^x = 37# #log_6 6^x=log_6 37# #loga^b=bloga# #xlog_6 6 =log_6 37# #log_a a=1# #x = log_6 37# 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 1423 views around the world You can reuse this answer Creative Commons License