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Does anyone know how to express log √(x^4 / yz) in terms of sums and differences of logarithms

2007-04-22 15:32:53 · 3 answers · asked by Anonymous in Science & Mathematics Mathematics

3 answers

Yes. Remember your log rules:
log (xy) = log x + log y
log (x/y) = log x - log y
log (x^a) = a log x

Also, √x = x^(1/2), so we get
log √x = log (x^(1/2)) = (1/2) log x

So, log √(x^4 / yz)
= log ((x^4 / yz)^(1/2))
= (1/2) log (x^4 / yz)
= (1/2) (log (x^4) - log (yz))
= (1/2) (4 log x - (log y + log z))
= 2 log x - (1/2) log y - (1/2) log z.

You could also choose to evaluate the square root before the log, and write everything as a multiplication: you'd have
log √(x^4 / yz)
= log ((x^4.y^(-1).z^(-1))^(1/2))
= log (x^2.y^(-1/2).z^(-1/2))
= log (x^2) + log (y^(-1/2)) + log (z^(-1/2))
= 2 log x - (1/2) log y - (1/2) log z
as before.

2007-04-22 15:36:02 · answer #1 · answered by Scarlet Manuka 7 · 0 0

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2016-05-21 04:06:45 · answer #2 · answered by garnet 3 · 0 0

log of a product is the sum of the logs; log of a fraction is log of numerator minus log of denominator. log of a power is the power times the log.

log√[x^4/yz] = 0.5*log(x^4/yz) = 0.5*[log(x^4) - log(yz)] = 0.5*[4*log(x) - log(y) - log(z)] = 2*log(x) - 0.5*log(y) - 0.5*log(z)

2007-04-22 15:41:36 · answer #3 · answered by gp4rts 7 · 0 0

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