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anonymous
 one year ago
Precalc Logarithm help?
anonymous
 one year ago
Precalc Logarithm help?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Evaluate each of the following

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0@amistre64 @sammietaygreen @zepdrix

sweetburger
 one year ago
Best ResponseYou've already chosen the best response.0these all use the properties of logarithms do you know what these are?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I honestly don't understand the first thing about logarithms

freckles
 one year ago
Best ResponseYou've already chosen the best response.0\[\ln(\frac{a}{b})=\ln(a)\ln(b) \text{ Try applying the quotient rule \to the first one } \\ \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.01e^5? I don't think I understand....

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0which one you stuck on?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0they 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
 one year ago
Best ResponseYou've already chosen the best response.0for 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
 one year ago
Best ResponseYou've already chosen the best response.0second one is even easier \[\large 10^{\log_{10}(\text{whatever})}=\text{whatever}\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0third one is similar the bases are the same so \[\log_{2015}(2015^{1})=?\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0all 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
 one year ago
Best ResponseYou've already chosen the best response.0also 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
 one year ago
Best ResponseYou've already chosen the best response.0because you still have to use the inverse thingy

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0and all this time i thought the quotient rule was from calculus...

freckles
 one year ago
Best ResponseYou've already chosen the best response.0they call a lot of things the quotient rule
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