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 1734 views around the world You can reuse this answer Creative Commons License