How do solve #2.6^x = 20.4#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer De Rono Nov 2, 2015 #x =log_(2.6) 20.4 # Explanation: #2.6^x= 20.4# #implies log_(2.6) 2.6^x =log_(2.6) 20.4# #implies x =log_(2.6) 20.4 # 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 1512 views around the world You can reuse this answer Creative Commons License