Question

Help please!!!!!!!!!!!!​

Answer 1

`\log_(9){(13)/(x)}=\log_(9){\boxed{13}}-log_9(x)`

Explanation:

The rule is ...

log(a/b) = log(a) -log(b) . . . . all logs to the same base

Then for a/b = 13/x, this becomes ...

log(13/x) = log(13) -log(x)

If the base of logarithms is 9, then this is ...

log9(13/x) = log9(13) -log9(x)