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August 28, 2014

August 28, 2014

Posted by **Jus** on Monday, May 25, 2009 at 5:38pm.

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

- Math -
**Reiny**, Monday, May 25, 2009 at 5:55pm1.

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.

- Math -
**Jus**, Monday, May 25, 2009 at 6:18pmthank 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

- Math -
**Reiny**, Monday, May 25, 2009 at 6:29pmone 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

- Math -
**Nora Escobedo**, Tuesday, October 11, 2011 at 2:01amWhich expression eqivalent to 27x^-2y^6/3x^5y^2z^9?

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