Question

logs

Answer 1

Dance dance dance

Answer 2

bricks

Answer 3

aluminium bars

Answer 4

\[\log_{9} (1/3)\]

Answer 5

hahah yall are to funny!

Answer 6

Are you allowed to use a calculator?

Answer 7

yess

Answer 8

Familiar with change of base formula?

Answer 9

nopeee

Answer 10

zepp formula will always work and good if you are trying to get and answer quick and not worry too much
but don't forget that
\[\log_b(x)=y\iff b^y=x\]and you should be able to switch back and forth quicklyi

Answer 11

\[\log_9(\frac{1}{3})=y\iff 9^y=\frac{1}{3}\]
if you can solve this you are in good shape
if not, you need to get familiar with exponents

Answer 12

you should be thinking "9 to what power gives \(\frac{1}{3}\) ?"

Answer 13

yes i can get myself to this point.. I just need to get familiar with exponents.. ha but really i need help with the last part you said

Answer 14

ok as long as you recognize that is the crux of the problem
now we can think, how can i get from 9 to \(\frac{1}{3}\) only using EXPONENTS
i.e not multiplying or dividing or anything like that

Answer 15

hint

\(\sqrt{x}=x^\frac{1}{2}\)

Answer 16

first off it is pretty clear that the square root of 9 is 3 right? now we recall that we can write \(\sqrt{9}=3\) or
\(9^{\frac{1}{2}}=3\)

Answer 17

And forget the fraction

9 to the power of what gives ^3

Answer 18

3?

Answer 19

that gets us the 3, now we need the reciprocal

Answer 20

3/1

Answer 21

Reciprocal of the exponent

Answer 22

opps 1/3

Answer 23

\[\large 9^\frac{1}{2}=\sqrt{9}=3\]

Answer 24

Reciprocal of the exponent, not the result D:

Answer 25

We have to do something so 3 becomes \(\frac{1}{3}\)

Answer 26

okay.. im getting confused here 2/1? idk

Answer 27

Therefore

-1/2

Answer 28

sorry i just get confused with all these terms :/

Answer 29

might i make a suggestion? because it looks like you are confused as to where to focus
we have
\[\log_b(x)=y\iff b^y=x\] so for example
\[10^2=100\iff \log_{10}(100)=2\]
\[5^3=125\iff \log_5(125)=3\]
\[10^{-3}=0.001\iff \log_{10}(0.001)=-3\]
\[36^{\frac{1}{2}}=6\iff\log_{36}(6)=\frac{1}{2}\]
\[4^{-1}=\frac{1}{4}\iff \log_4(\frac{1}{4})=-1\] and so one
what you need to focus on is the exponent

Answer 30

therefore when you see
\[\log_9(\frac{1}{3})=y\] you think
\[9^y=\frac{1}{3}\]
a minus sign with give you the reciprocal and the one half will give you the square root, so your answer is minus one half

Answer 31

I wish I can explain that well @satellite73 :P

Answer 32

YAY! no @zepp you can explain very well i just get confused because everything was spread all over the place.. but what i do want to learn is how you got -1/2?

Answer 33

:o

Thanks :)

And for the reciprocal thing, it's very easy

Here \[4^1=4\] right?

Answer 34

yesss

Answer 35

Alright! What you have to remember that if the exponent is negative, it will flip the number(or fraction)

\[4^-1=\frac{1}{4}\]

Answer 36

\[4^{-1}=\frac{1}{4}\] sorry

Answer 37

its okay, i understood ha

Answer 38

Great! :D

░░░░░░░░░░░░▄▄
░░░░░░░░░░░█░░█
░░░░░░░░░░░█░░█
░░░░░░░░░░█░░░█
░░░░░░░░░█░░░░█
███████▄▄█░░░░░██████▄
▓▓▓▓▓▓█░░░░░░░░░░░░░░█
▓▓▓▓▓▓█░░░░░░░░░░░░░░█
▓▓▓▓▓▓█░░░░░░░░░░░░░░█
▓▓▓▓▓▓█░░░░░░░░░░░░░░█
▓▓▓▓▓▓█░░░░░░░░░░░░░░█
▓▓▓▓▓▓█████░░░░░░░░░█
██████▀░░░░▀▀██████▀

Answer 39

how did you do that????

Answer 40

hax

Answer 41

but to get 1/2 in general.. was it just guess and check? or what

Answer 42

Haha http://fb-art.blogspot.ca/2011/08/big-cool-facebook-like-thumb-special.html there you go :P

Answer 43

Nope, since the square root of 9 is 3, squareroot also means ^1/2, and we want the 3 flipped, so we add a negative sign, -1/2

Answer 44

in a way it is a guess, but an educated guess
you have 9 as the base
and you see a 3
you get from 9 to 3 by taking the square root, and you can write the square root as an exponent

Answer 45

of course you can also get from 9 to 3 by
a) subtracting 6
b) dividing by 3
c) etc etc
but you want something that you can write as an exponent, like taking powers, roots, and reciprocals

Answer 46

so i would say "educated guess" and practice with exponents

Answer 47

THANK YOU BOTH! YAY! i understand it now!!!

Answer 48

..oh forsure! ha