Answer:
[tex]\log_{9}{\dfrac{13}{x}}=\log_{9}{\boxed{13}}-\log_{9}{x}[/tex]
Step-by-step 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)
