How do you solve 2^x=52x=5? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer P dilip_k May 6, 2016 x=log5/log2~~2.32x=log5log2≈2.32 Explanation: Given' 2^x=52x=5 Taking log of both sides =>log2^x=log5⇒log2x=log5 =>xlog2=log5⇒xlog2=log5 =>x=log5/log2=0.6989/0.3010~~2.32⇒x=log5log2=0.69890.3010≈2.32 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 26553 views around the world You can reuse this answer Creative Commons License