looks good to me. I think
log_2(128(x-5)) is better that "log of 2", since most people would read that as just "log 2" which is not what you meant.
Actually, I kind of like 7+log_2(x-5) but either is fine.
You also need to specify that the domain is
(x-5) > 0 or x > 5
find the equation of the inverse of f(x)= 2^x-7 + 5
I did x=2^y-7+ 5
-5 -5
x-5=(2^y)(2^-7)
128*x-5=2^y
answer y=log of 2 (128(x-5))
I switched x and y and solved for y
Is this correct? Thanks for checking my work.
2 answers
Yes , I wasn't sure the proper way of writing the log with the number under it. Thanks for explaining how to do it, so I will know in the future. Thank you for checking my work.