The Attempt at a Solution
would that equal m^9 or m^(8*√5)
and thanks for your help with the first one :) really appreciate it
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????
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?