How do you solve #log_2x=log_4 25#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Konstantinos Michailidis Mar 15, 2016 It is #log_2 x=log_(2^2) 25=>(logx/log2) = (log25/log2^2)=> logx/log2=(2log5)/(2log2)=>logx=log5=>x=5# 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 1072 views around the world You can reuse this answer Creative Commons License