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Algebra
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OpenStudy (anonymous):
The problem: log(x^(3)y^(2))-2logx√3:(y)-3log((x)/(y))
My answer: log((y^((14)/(3)))/(x^(2)))
Book's answer: log((y^((13)/(3)))/(x^(2)))
What am I doing wrong?
13 years ago
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OpenStudy (anonymous):
i am a bit confused as to the original problem
it is
\[\log(x^2y^2)-2\log(x\sqrt[3]{y})-3\log(\frac{x}{y})\]?
13 years ago
OpenStudy (anonymous):
Almost. The first x^{3}
13 years ago
OpenStudy (anonymous):
is not x^{2}
13 years ago
OpenStudy (anonymous):
\[\log(x^3y^2)-2\log(x\sqrt[3]{y})-3\log(\frac{x}{y})\]
13 years ago
OpenStudy (anonymous):
yes
13 years ago
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OpenStudy (anonymous):
and you want to combine in to a single log right?
13 years ago
OpenStudy (anonymous):
yes
13 years ago
OpenStudy (anonymous):
ok then it is mostly bookkeeping with the exponents
13 years ago
OpenStudy (anonymous):
\[\log(x^3y^2)-2\log(x\sqrt[3]{y})-3\log(\frac{x}{y})\]
\[\log(x^3y^2)-\log(x^2\sqrt[3]{y^2})-\log(\frac{x^3}{y^3})\]
13 years ago
OpenStudy (anonymous):
the input is
\[\frac{x^3y^2y^3}{x^2y^{\frac{2}{3}}x^3}\]
13 years ago
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OpenStudy (anonymous):
simplifies to
\[\frac{y^{\frac{13}{3}}}{x^2}\]
13 years ago
OpenStudy (anonymous):
oh that is what you got!
i think your answer is right
13 years ago
OpenStudy (anonymous):
Ok now I see where my mistake was. I didn't distribute the 2 in front of the log, only applied it to the x.
13 years ago
OpenStudy (anonymous):
I was off and I couldn't figure it out why. Now i know.
13 years ago
OpenStudy (anonymous):
oh that is the book answer
ok i hope it is clear
13 years ago
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OpenStudy (anonymous):
yes, thank you very much for your help
13 years ago
OpenStudy (anonymous):
yw
13 years ago
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