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kwarr5 · Mar 18, 2011, 01:39 PM

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

HELP!

ebaines · Mar 18, 2011, 02:52 PM

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?

kwarr5 · Mar 18, 2011, 03:17 PM

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

kwarr5 · Mar 18, 2011, 03:17 PM

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

Unknown008 · Mar 18, 2011, 11:57 PM

$\log_5 7$

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?