How do you solve #5^x = 10#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Alan P. Feb 24, 2016 #x=log_5 10 ~~1.430677# Explanation: Make use of the standard relationship #color(white)("XXX")log_b a = c <=> b^c = a# or (reversed) #color(white)("XXX")b^c=a <=> log_b a# So #color(white)("XXX")5^x= 10 <=> log_5 10 = x# #log_5 10# can be evaluated using a calculator as #~~1.430677# 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 3147 views around the world You can reuse this answer Creative Commons License