Solve for x: $$\log_{6}(x+4)-\log_{6}(x-4)=\log_{6}3 $$

Question

Solve for x:  $$\log_{6}(x+4)-\log_{6}(x-4)=\log_{6}3   $$

anonymous · Answer

@satellite73  @phi @amistre64

amistre64 · Answer

$$log~a-log~b=log~\frac ab$$ should be useful

amistre64 · Answer

also:$$if~log_n(p)=log_n(q)~then~p=q$$

anonymous · Answer

so it looks like $$\log_{6}\frac{ x+4 }{ x-4 } =\log_{6}3 $$

anonymous · Answer

as the first step..

phi · Answer

yes, now take the anti-log of both sides
then multiply both sides by (x-4)

phi · Answer

take the anti-log means  drop the logs
or, mathematically,  make each side the exponent of 6  and 6 ^ (log_6(stuff)) = stuff

anonymous · Answer

i got 8 as my final answer..  is this right?

zepdrix · Answer

yah that sounds right, good job!

anonymous · Answer

ok cool so you got the same answer then?

zepdrix · Answer

yah

anonymous · Answer

great!  i feel better then haha

phi · Answer

It is easy to check your answer.  Start with the original question
$$ \log_{6}\frac{ x+4 }{ x-4 } =\log_{6}3 $$
replace x with 8
$$ \log_{6}\frac{ 8+4 }{ 8-4 } =\log_{6}3 $$
simplify
$$ \log_{6}\frac{ 12 }{ 4 } =\log_{6}3 $$
$$ \log_{6}3 =\log_{6}3 $$
which looks good...