anonymous
  • anonymous
using the Law of Logarithms to expand the expression.. Please help!
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
\[\log \left( \frac{ 10^{x} }{ x(x ^{2}+1)(x ^{4}+2) } \right)\]
freckles
  • freckles
have you considered first using the quotient rule for log? \[\log(\frac{u}{v})=\log(u)-\log(v) \\ \]
freckles
  • freckles
\[\log(uv)=\log(u)+\log(v) \text{ is another rule you can use after that } \\ \log(u^r)=rlog(u) \text{ is the power rule which can also be used here }\]

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anonymous
  • anonymous
\[\log10^{x}-logx (x ^{2}+1)+\log(x ^{4}+2)\]
freckles
  • freckles
I think you forgot to distribute the minus sign
anonymous
  • anonymous
\[xlog10-logx(x ^{2}+1)+\log(x ^{4}+2)\]
anonymous
  • anonymous
but since the denominator is looks like multiplication wouldn't be a addition?
freckles
  • freckles
\[\log(\frac{10^x}{x(x^2+1)(x^4+2)}) \\ \log(10^x)-\log(x(x^2+1)(x^4+2)) \\ \log(10^x)-[\log(x(x^2+1)(x^4+2)) ] \\ \log(10^x)-[\log(x(x^2+1))+\log(x^4+2)]] \\ \log(10^x)-\log(x(x^2+1))-\log(x^4+2) \]
freckles
  • freckles
though could also expand the log(x(x^2+1)) part too
anonymous
  • anonymous
ooh so the signs changed because of the minus sign before the brackets?
freckles
  • freckles
yep that is called the distributive property
anonymous
  • anonymous
oh ok.. sorry i suck on paying attention to small details.
anonymous
  • anonymous
ok so could i expand more after log(x(x^2+1))−log(x^4+2)?
freckles
  • freckles
the log(x(x^2+1)) could be expanded more
freckles
  • freckles
notice x(x^2+1) is a product of x and (x^2+1)
anonymous
  • anonymous
would i t be: Logx+log (x^2+1)
freckles
  • freckles
yes and don't forget we also have a - sign in front of the log(x(x^2+1) so you have: \[\log(\frac{10^x}{x(x^2+1)(x^4+2)}) \\ \log(10^x)-\log(x(x^2+1)(x^4+2)) \\ \log(10^x)-[\log(x(x^2+1)(x^4+2)) ] \\ \log(10^x)-[\log(x(x^2+1))+\log(x^4+2)]] \\ \log(10^x)-\log(x(x^2+1))-\log(x^4+2) \\ \log(10^x)-[\log(x)+\log(x^2+1)]-\log(x^4+2)\]
freckles
  • freckles
anyways yes log(10^x) =xlog(10) but assuming this is log base 10 you could also simplify the x log(10) more
anonymous
  • anonymous
it will just become x right?
freckles
  • freckles
right
anonymous
  • anonymous
since log 10=1
freckles
  • freckles
yes
anonymous
  • anonymous
the result is: \[x-\log (x)-\log (x ^{2}+1)-\log (x ^{4}-2)\]
anonymous
  • anonymous
?
freckles
  • freckles
well you change the sign between x^4 and 2 for some reason
anonymous
  • anonymous
oh haha yeah.. oops
freckles
  • freckles
\[\log(\frac{10^x}{x(x^2+1)(x^4+2)}) \\ \log(10^x)-\log(x(x^2+1)(x^4+2)) \\ \log(10^x)-[\log(x(x^2+1)(x^4+2)) ] \\ \log(10^x)-[\log(x(x^2+1))+\log(x^4+2)]] \\ \log(10^x)-\log(x(x^2+1))-\log(x^4+2) \\ \log(10^x)-[\log(x)+\log(x^2+1)]-\log(x^4+2) \\ x \log(10)-\log(x)-\log(x^2+1)-\log(x^4+2) \\ x(1)-\log(x)-\log(x^2+1)-\log(x^4+2) \\ x-\log(x)-\log(x^2+1)-\log(x^4+2)\] should be right if the log's are in base 10

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