How do you solve #5(2^y)=3-(2^(y+2))#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Cesareo R. May 29, 2016 #y =-log_2 3 # Explanation: #5times 2^y-3+4times2^y=0->9times 2^y-3=0#. Applying logarithm to both sides #y log_e 2=log_e(1/3)#. Solving for #y# #y = (log_e(1/3))/(log_e 2) = -log_2 3 # 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 879 views around the world You can reuse this answer Creative Commons License