anonymous
  • anonymous
Precalc Logarithm help?
Mathematics
katieb
  • katieb
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anonymous
  • anonymous
i'll help you
anonymous
  • anonymous
Evaluate each of the following
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anonymous
  • anonymous

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sweetburger
  • sweetburger
these all use the properties of logarithms do you know what these are?
anonymous
  • anonymous
I honestly don't understand the first thing about logarithms
anonymous
  • anonymous
freckles
  • freckles
\[\ln(\frac{a}{b})=\ln(a)-\ln(b) \text{ Try applying the quotient rule \to the first one } \\ \]
anonymous
  • anonymous
1-e^5? I don't think I understand....
anonymous
  • anonymous
which one you stuck on?
anonymous
  • anonymous
they are all the same really they are all using the same two facts \[\huge \log_b(b^x)=x\] and \[\huge b^{\log_b(x)}=x\] because the log is the inverse of the exponential
anonymous
  • anonymous
for the first one \[\ln(\frac{1}{e^5})\] rewrite as \[\ln(e^{-5})\] then use the fact that \[\ln(x)=\log_e(x)\] so you have \[\log_e(e^{-5})=-5\]
anonymous
  • anonymous
second one is even easier \[\large 10^{\log_{10}(\text{whatever})}=\text{whatever}\]
anonymous
  • anonymous
third one is similar the bases are the same so \[\log_{2015}(2015^{-1})=?\]
anonymous
  • anonymous
all of these, once written in exponential form, are kind of like "who's buried in grant's tomb"? what power would you raise 2015 to in order to get \(2015^{-1}\)? i guess \(-1\)
freckles
  • freckles
also applying the quotient rule to the first one looks like this: \[\ln(\frac{1}{e^5})=\ln(1)-\ln(e^5) \\ \text{ then use that } \ln(1)=0 \\ \text{ so } \ln(\frac{1}{e^5})=-\ln(e^5) \\ \text{ but you could have gotten away from using that rule as } satellite \text{ said }\]
freckles
  • freckles
because you still have to use the inverse thingy
anonymous
  • anonymous
and all this time i thought the quotient rule was from calculus...
anonymous
  • anonymous
inverse thingy lol
freckles
  • freckles
they call a lot of things the quotient rule

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