The Attempt at a Solution
would that equal m^9 or m^(8*√5)
No problem.and thanks for your help with the first one :) really appreciate it
Use the base of 6. On the leftside, you have 6^(-x). On the rightside you have (6^2)^(x+3)=6^(2(x+3))right now im stuck on this problem (1/6)^x = 36^(x+3) if i solve it by making the bases equal to 36 by multiplying 1/6 by -2 i get -2x but then for x i get -3 which obviously would not work, but if i use -2 as x the equation works BUT how do i get -2 from that equation????
You don't need log identities, just the fact that the exponential functions (ax for any positive a) are "one-to-one"my alg2 teacher said inorder to solve (1/6)^x = 36^(x+3) you need to make the BASES equal to each other i can easily make 1/6 to 36 by raising it to -2 but then i have to multiply x by -2. How do the log identities work?