How do you solve Log_2 x + log_4 x + log_8 x = 11 log2x+log4x+log8x=11? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Shwetank Mauria · EZ as pi Mar 8, 2018 x=64x=64 Explanation: As log_ab=logb/logalogab=logbloga, we can write log_2x+log_4x+log_8x=11log2x+log4x+log8x=11 as logx/log2+logx/log4+logx/log8=11logxlog2+logxlog4+logxlog8=11 or logx/log2+logx/(2log2)+logx/(3log2)=11logxlog2+logx2log2+logx3log2=11 or logx/log2(1+1/2+1/3)=11logxlog2(1+12+13)=11 or logx/log2((6+3+2)/6)=11logxlog2(6+3+26)=11 or logx/log2 (11/6)=11logxlog2(116)=11 or logx/log2=11xx6/11=6logxlog2=11×611=6 i.e. logx=6log2=log2^6logx=6log2=log26 and x=2^6=64x=26=64 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)=819x−4=81? How do you solve logx+log(x+15)=2logx+log(x+15)=2? How do you solve the equation 2 log4(x + 7)-log4(16) = 22log4(x+7)−log4(16)=2? How do you solve 2 log x^4 = 162logx4=16? How do you solve 2+log_3(2x+5)-log_3x=42+log3(2x+5)−log3x=4? See all questions in Logarithmic Models Impact of this question 32258 views around the world You can reuse this answer Creative Commons License