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log C = log P + µ log V

thanks

2007-11-26 06:56:37 · 6 answers · asked by Anonymous in Science & Mathematics Mathematics

6 answers

C = PV^(μ)

2007-11-26 07:01:37 · answer #1 · answered by Como 7 · 3 1

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 · answer #2 · answered by Puzzling 7 · 1 2

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 · answer #3 · answered by pink 2 · 0 1

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 · answer #4 · answered by Niaómh M 2 · 0 2

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 · answer #5 · answered by Anonymous · 0 0

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 · answer #6 · answered by fcas80 7 · 2 2

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