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## anonymous one year ago Precalc Logarithm help?

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1. anonymous

i'll help you

2. anonymous

Evaluate each of the following

3. anonymous

@amistre64 @sammietaygreen @zepdrix

4. sweetburger

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

5. anonymous

I honestly don't understand the first thing about logarithms

6. anonymous

@sweetburger

7. freckles

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

8. anonymous

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

9. anonymous

which one you stuck on?

10. 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

11. 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 $\log_e(e^{-5})=-5$

12. anonymous

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

13. anonymous

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

14. 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 $$-1$$

15. freckles

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

16. freckles

because you still have to use the inverse thingy

17. anonymous

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

18. anonymous

inverse thingy lol

19. freckles

they call a lot of things the quotient rule

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