Posted by Jus on Monday, May 25, 2009 at 5:38pm.
1.
let a^x =m and a^y = n
by definition:
a^x = m <---> log a m = x and
a^y = n <---> log a n = y
just for the fun of it, I am going to add the two equations on the right ...
log a m + log a n = x+y
or log a mn = x+y, since we are supposed to know the product rule of logs
but that can be changed to
a^(x+y) = mn
but what was m and n ??
so (a^x)(a^y) = mn = a^(x+y)
which is "product law of exponents"
Q.E.D.
thank you
could you help me with part 3 too?
I know that part two is the same as part one except you divide. please help with part 3, thank you in advance reiny
one of the log rules states
log b a = loga/logb ...(base 10, or any other legal base you want to choose)
so again by definition of the log notation
a=b^(loga/logb) for any a,b>0
<-----> log b a = loga/logb
well, well, that was even easier than the first one.
let's test it
let a = 8 and b=2
we know log 2 8 = 3
and log8/log2 = .903089987/.301029995 = 3
Which expression eqivalent to 27x^-2y^6/3x^5y^2z^9?
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