Logs

The log of a quotient rule (in reverse):

\(\displaystyle \mbox{ \log_{10}{(x + 6)} - \log_{10}{(x - 6)} = \log_{10}{\left(\frac{x + 6}{x - 6}\right)}\)

The equation is now

\(\displaystyle \mbox{ \log_{10}{\left(\frac{x + 6}{x - 6}\right) = 1}\)
 
\(\displaystyle \L
\log _{10} \left( {\frac{{x + 6}}{{x - 6}}} \right) = 1\quad \Rightarrow \quad \left( {\frac{{x + 6}}{{x - 6}}} \right) = 10\)

You solve for x.
 
Christina82 said:
How do you find x from there? The answer is 22/3....
Click to expand...

Your previous helper got to

log10 = (x+6)/(x-6) and then said

log10 (x+6)/(x-6) = 1

Now ... recognize that log{sub]mx = y means

m^y = x

Use that to rewrite your problem without the "log", and then you can solve it.
 
\(\displaystyle \mbox{ \frac{x + 6}{x - 6} = 10}\)

Multiply both sides by \(\displaystyle \mbox{(x - 6)}\):

\(\displaystyle \mbox{ x + 6 = 10(x - 6)}\)
 
Are you really telling us that you cannot solve \(\displaystyle \frac{{x + 6}}{{x - 6}} = 10\)???
 
Oh, I must not have been looking at it correctly. Thanks for clarifying!!

pka, that wasn't a very nice comment... I'm capable of solving it, it was just a mistake. :wink:
 
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