How do you solve log_2 x+log_8 32x=1? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer P dilip_k ยท ali ergin Apr 23, 2016 log_2x+log_8 32x=1 =>log_2x+log_8 32+log_8x=1 =>log_2x+log_8 2^5+log_8x=1 =>log_2x+5log_8 2+log_8x=1 =>log_2x+5xx1/log_2 8+1/log_x 8=1 =>log_2x+5xx1/log_2 2^3+1/log_x 2^3=1 =>log_2x+5xx1/(3log_2 2)+1/(3log_x 2)=1 =>log_2x+5/3+1/3xxlog_2 x=1 =>3log_2x+5+log_2 x=3 =>4log_2x=3-5=-2 =>log_2 x=-1/2 =>x=2^(-1/2)=1/sqrt2 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 1862 views around the world You can reuse this answer Creative Commons License
