Question

Anyone good with logarithms?

Answer 1

\[\log_{8} (n-3)+\log_{8}( n+4)=1\]

Answer 2

use the additive rule of logs to combine the first two terms

log8( (n-3)(n+4)=1
converting back to exponential
8^1=(n-3)(n+4) foiling the right side now you get n^2+4n-3n-12

move the 8 over to to put into quadratic form
n^2+n-20=0

you can't factor this normally, so use the quadratic formula and solve for n