How do you solve #2^(x+2) - 2^(x+5)= -7#? Precalculus Solving Exponential and Logarithmic Equations Logarithmic Models 1 Answer Konstantinos Michailidis Sep 9, 2015 We have that #2^(x+2)-2^(x+5)=-7=>2^x(2^2-2^5)=-7=>2^x*(-28)=-7=>2^x=7/28=>2^x=1/4=>2^x=2^(-2)=>x=-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 1671 views around the world You can reuse this answer Creative Commons License