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.
1. Given the product law of logarithms, prove the product law of exponents.
2. Given the quotient law of logarithms, prove the quotient law of exponents.
3. Apply algebraic reasoning to show that
a=b^(loga/logb) for any a,b>0
Please explain these to me.
All I know is that
The product law of logs are:
Log(AB)=logA+logB
The Quotient law of logs are:
Log(A/B)=logA-Logb
5 answers
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
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
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?
Dr Disrespect
2 answers