To start with, log(X/Y) = log X - log Y.
So the first term becomes log a - log b, the second one (taking the negative into account) becomes -log c + log b, etc. We wind up with:
log a - log b - log c + log b + log d - log c - 2 log a + 2 log c + log a - log 10 - log d.
You'll notice that the log a's, the log b's, the log c's, and the log d's all cancel, leaving -log 10. Log 10 is 1, so the whole thing simplifies to -1.
Another way to do it would be to combine all the logs into one log of a complex fraction, which would them simplify to 1/10, but that's too hard to show on Yahoo Answers. :)
Hope that helps!
2006-09-29 09:37:43
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answer #1
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answered by Jay H 5
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Others have already given you the answer, but I prefer to show askers how to solve the problems themselves. There are three basic properties of logarithms:
log (x/y) = log x - log y
log (xy) = log x + log y
(Also, 2 log x = log (x^2), but that isn't relevant here)
Using these properties, you can expand the expression into a sum/difference of simple logs which is then easy to simplify.
2006-09-29 17:50:21
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answer #2
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answered by kslnet 3
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= log(a/b / c/b) + log(d/c) -log((a/c)^2) + log(a/10d)
= log(a/b / c/b) + log(d/c / ((a/c)^2)) + log(a/10d)
= log ((a/b / c/b).(d/c / ((a/c)^2)) + log(a/10d)
= log ((ad/bc).(bc^2/ca^2) + log(a/10d)
= log(adbc/bca^2) + log(a/10d)
= log(ad/a^2) + log(a/10d)
= log((ad/a^2).(a/10d)
= log(da^2/10da^2) = log(1/10)
2006-09-29 16:50:26
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answer #3
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answered by daniel_cohadier 3
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log(a/b) - log(c/b) + log(d/c) - 2log(a/c) + log(a/(10d))
log(a/b) - log(c/b) + log(d/c) - log((a/c)^2) + log(a/(10d))
log(((a/b) * (d/c) * (a/(10d)))/((c/b) * (a/c)^2))
log(((da^2)/(10bcd))/((ca^2)/(bc^2)))
log(((a^2)/(10bc)) / ((a^2)/(bc)))
log(((a^2)/(10bc)) * ((bc)/(a^2)))
log((bca^2)/(10bca^2))
log(1/10)
ANS : -1
2006-09-29 17:22:06
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answer #4
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answered by Sherman81 6
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yikes
2006-09-29 16:32:59
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answer #5
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answered by Anonymous
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log((a/b)/(c/b)*(d/c)/(a^2/c^2)*(a/10d))
log( abdc^2a/bcca^210d)
log(a^2c^2bd/bc^2a^2d10)
log1/10)=-log(10)
2006-10-02 23:58:23
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answer #6
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answered by yupchagee 7
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