How do you solve #.5(6^x) = 2^x#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer moutar Jan 13, 2016 #x=0.631# (3sf) Explanation: Take logs of both sides. #log(0.5*(6^x))=log(2^x)# #log0.5 + xlog6 = xlog2# #log0.5=xlog2-xlog6# #log0.5=x(log2-log6)# #x=(log0.5)/(log2-log6)# #x=log2/(log6-log2)# #x=log2/log3# #x=0.631# (3sf) 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 1582 views around the world You can reuse this answer Creative Commons License