- #1
davedave
- 50
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This is not a homework question. I just try it for enjoyment.
Let L = log to the base x of (yz) M = log to the base y of (xz) and
N = log to the base z of (xy)
This is how I do it without much luck.
I put all the equations in exponential form
yz = x^L xz = y^M xy = z^N
raise the right-hand sides of the equations to the required power so that x y z will have a product of LMN in the exponent.
(yz)^(MN) = x^(LMN) (xz)^(LN) = y^(LMN) (xy)^(LM) = z^(LMN)
combining the equations gives (yz)^(MN) * (xz)^(LN) * (xy)^(LM) = (xyz)^(LMN)
now, multiply both sides of the equation to (xyz)^-2 and rearrange the terms
x^(L(M+N)-2) * y^(M(L+N)-2) * z^(N(L+M)-2) = (xyz)^(LMN-2)
But, if I could make the left side of the equation above into (xyz)^(L+M+N), then that would complete the proof.
Can someone think of any other way of doing this proof? Thanks.
sorry about my omission.
Here is the question.
Prove that L + M + N = LMN - 2
Let L = log to the base x of (yz) M = log to the base y of (xz) and
N = log to the base z of (xy)
This is how I do it without much luck.
I put all the equations in exponential form
yz = x^L xz = y^M xy = z^N
raise the right-hand sides of the equations to the required power so that x y z will have a product of LMN in the exponent.
(yz)^(MN) = x^(LMN) (xz)^(LN) = y^(LMN) (xy)^(LM) = z^(LMN)
combining the equations gives (yz)^(MN) * (xz)^(LN) * (xy)^(LM) = (xyz)^(LMN)
now, multiply both sides of the equation to (xyz)^-2 and rearrange the terms
x^(L(M+N)-2) * y^(M(L+N)-2) * z^(N(L+M)-2) = (xyz)^(LMN-2)
But, if I could make the left side of the equation above into (xyz)^(L+M+N), then that would complete the proof.
Can someone think of any other way of doing this proof? Thanks.
sorry about my omission.
Here is the question.
Prove that L + M + N = LMN - 2
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