How do you solve #2(5^(6x))-9(5^(4x))+13(5^(2x))-6 = 0#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer dani83 Aug 14, 2015 # x = 0 " or " x = 1/2log_5 2" or " x = 1/2log_5 (3/2) # Explanation: # 2(5^(6x))-9(5^(4x))+13(5^(2x))-6 = 0 # Take # y = 5^(2x) # # 2y^3-9y^2+13y-6 = 0 # # (y-1)(y-2)(2y-3) = 0 # # 5^(2x) = 1 " or " 5^(2x) = 2" or " 5^(2x) = 3/2 # # x = 0 " or " x = 1/2log_5 2" or " x = 1/2log_5 (3/2) # 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 2157 views around the world You can reuse this answer Creative Commons License