C = PV^(μ)
2007-11-26 07:01:37
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answer #1
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answered by Como 7
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Use the exponent rule first:
µ log V = log V^µ
Next use the product/addition rule of logs:
log a + log b = log(ab)
log P + log V^µ
= log( P * V^µ )
Putting it all together:
log C = log( P * V^µ )
Raise both sides upon the power 10:
C = P * V^µ
Note: I used the symbol * to mean multiplication (so it wouldn't be confused with x as a variable). In this case you could just push your terms together as shown:
C = PV^µ
Don't confuse this with:
C = (PV)^µ
The answer is:
C = P(V^µ)
2007-11-26 15:03:17
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answer #2
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answered by Puzzling 7
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log C = log P + µ log V
as from the logarithmic identity, b*log(a)=log(a^b),so,
logC=logP+log(V^µ )
as from the logarithmic identity, logAB=logA+logB
logC=logP(V^µ )
C=P(V^µ )
2007-11-26 19:28:19
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answer #3
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answered by pink 2
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i wont do it for you, but remember, logs are the opposite of normal functions. put it into composite or invve a bash at it. if all else fails, go to asktheteacher.com
good luck :)
2007-11-26 15:01:12
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answer #4
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answered by Niaómh M 2
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i dont no but if you get a answer from nats so hot plz report it becouse this is what she does
ghslh
2007-11-26 14:59:39
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answer #5
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answered by Anonymous
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log C = log P + µ log V
log C = log P + log (V^µ)
log C = log [P * (V^µ)]
e^log C = e^log [P * (V^µ)]
C = [P * (V^µ)]
2007-11-26 15:02:55
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answer #6
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answered by fcas80 7
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