Question

Help please!!!!!!!!!!!!​

Answer 1

Answer:

  $\log_{9}{\dfrac{13}{x}}=\log_{9}{\boxed{13}}-\log_{9}{x}$

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)