Re: LOG

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Re: LOG

Not almost - EXACTLY,

Just remember that

if b^x = a, then x = log(a) / log(b)
because then
log(b^x) = log(a)
x * log(b) = log(a) {Why?}
etc.

regardless of the base of the logarithm as long as the base is
the same for the log in the numerator and the log in the denominator in the
first expression above.  So

log7(4) = log(4) / log(7) {also} = ln(4) / ln(7) {also} = log2(4) / log2(7).

In fact "log(a) / log(b)" is the inverse of "^", the exponentiation operator
because:

if b^x = a then x = log(a) / log(b) just like
if b+x = a then x =   a    -   b

Hope that helps.

-----

In article <1998033004242300.XAA04965@ladder03.news.aol.com>,
  ethan04@aol.com (ETHAN04) wrote:
>
> Can problems with logs be solved directly?
> Like   log7  4  =  2 log7  5 + log7 x+2
> Or almost?
>
> Thanks, Tom
>


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