How do you solve #27^(x+2)=3#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Don't Memorise May 23, 2015 #27^(x+2)=3# #27# can be written as #(3)^3# the expression now becomes: #(3)^(3.(x+2)) = (3)^1# equating exponents: #3.(x+2) = 1# #3x+6 = 1# #x= -5/3# 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 1628 views around the world You can reuse this answer Creative Commons License