Question

i don't know how to do this problem. i keep getting stuck. can anyone help me?
Solve:
log(base8)^(n-3) + log(base8)^n+4)= 1

Answer 1

Do you mean
log(base8)(n-3) + log(base8)(n+4)= 1 ?

then
log(base8)[(n-3)(n+4)[ = 1
(n-3)(x+4) = 8^1
n^2 + n - 12 - 8 = 0
n^2 + n - 20 = 0
(n+5)(n-4) = 0
n = -5 or n=4
but n=-5 would make log(base8)(--n+4) undefined,
so n = 4