Question

Precalc Logarithm help?

Answer 1

i'll help you

Answer 2

Evaluate each of the following

Answer 3

@amistre64 @sammietaygreen @zepdrix

Answer 4

these all use the properties of logarithms do you know what these are?

Answer 5

I honestly don't understand the first thing about logarithms

Answer 6

@sweetburger

Answer 7

\[\ln(\frac{a}{b})=\ln(a)-\ln(b) \text{ Try applying the quotient rule \to the first one } \\ \]

Answer 8

1-e^5? I don't think I understand....

Answer 9

which one you stuck on?

Answer 10

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

Answer 11

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 \[\log_e(e^{-5})=-5\]

Answer 12

second one is even easier \[\large 10^{\log_{10}(\text{whatever})}=\text{whatever}\]

Answer 13

third one is similar
the bases are the same so \[\log_{2015}(2015^{-1})=?\]

Answer 14

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 \(-1\)

Answer 15

also applying the quotient rule to the first one looks like this:
\[\ln(\frac{1}{e^5})=\ln(1)-\ln(e^5) \\ \text{ then use that } \ln(1)=0 \\ \text{ so } \ln(\frac{1}{e^5})=-\ln(e^5) \\ \text{ but you could have gotten away from using that rule as } satellite \text{ said }\]

Answer 16

because you still have to use the inverse thingy

Answer 17

and all this time i thought the quotient rule was from calculus...

Answer 18

inverse thingy lol

Answer 19

they call a lot of things the quotient rule