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
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?
Dr Disrespect
1. To prove the product law of exponents, we can start with two exponential expressions:
a^m * a^n
Applying the exponentiation property of multiplication, we can rewrite this expression as:
(a * a * a * ... * a) * (a * a * a * ... * a)
where the first multiplication has m factors of a, and the second multiplication has n factors of a.
Now, notice that we have a total of m + n factors of a. So we can write the above expression as:
a^(m + n)
Using the definition of exponents, this is equivalent to:
a^m * a^n = a^(m + n)
Therefore, we have proved the product law of exponents.
2. To prove the quotient law of exponents, we start with two exponential expressions:
(a^m) / (a^n)
Applying the exponentiation property of division, we can rewrite this expression as:
(a * a * a * ... * a) / (a * a * a * ... * a)
where the numerator has m factors of a, and the denominator has n factors of a.
Now, notice that in each term (a * a * a * ... * a), we have a total of m - n factors of a (because we subtract the corresponding factors in the denominator). So we can write the above expression as:
a^(m - n)
Using the definition of exponents, this is equivalent to:
(a^m) / (a^n) = a^(m - n)
Therefore, we have proved the quotient law of exponents.
3. To apply algebraic reasoning to show that a = b^(loga/logb) for any a, b > 0, we can start with the equation:
a = b^(loga/logb)
Take the logarithm of both sides of the equation using any base (let's use base 10 for simplicity):
log(a) = log(b^(loga/logb))
Apply the power rule of logarithms to the right side of the equation:
log(a) = (loga/logb) * log(b)
Notice that log(b) divided by log(b) is 1, so the equation becomes:
log(a) = log(a)
This is an identity, meaning it holds true for any value of log(a) and log(b).
Now, raise the base 10 to the power of both sides of the equation:
10^(log(a)) = 10^(log(a))
Using the definition of logarithms, we have:
a = a
Therefore, we have shown that a = b^(loga/logb) for any a, b > 0.