solve log5(x-4)=log7(x) for x.
5 and 7 are bases.

HELP!

To help get you started, use the rule for converting logarithm bases:


\log_a(b) = \frac {\log_c(b)} {\log_c(a)}


So to convert to a common logarithm base in the problemn you were given you can do this:


\log_7(x) = \frac {\log_5(x)}{\log_5(7)}


Now you can rewrite the original equation like this:


\log_5(x-4) = \frac {\log_5(x)}{\log_5(7)}\\


Can you take it from here?

well I get that.. I just don't understand where to take it from there the two x's are confusing me.

well I get that.. I just don't understand where to take it from there the two x's are confusing me

\log_5 7

Can be worked out. Then, you end up with something like:

\log\ x = a\log\ y

Raise a to become y to the power of a, to get:

\log_5 7

Can be worked out. Then, you end up with something like:

\log\ x = \log\ y^a

Now, can you continue?