Question

Evaluate log2 1 over 64.

Answer 1

−6

−32

6

32

Answer 2

Can you evaluate \[\log_2 64\]?

Answer 3

I'm not sure how, I'm willing to learn though :)

Answer 4

remember that \[\log_b n = a\]where \[b^a=n\]

Answer 5

so \[\log_2 64 = a \text{ such that } 2^a = 64\]

Answer 6

2*2=4
2*2*2=8
2*2*2*2=16
2*2*2*2*2=32
2*2*2*2*2*2=64

\[\log_2 64 = \]?

Answer 7

2^6?

Answer 8

well, 2^6= 64, yes. but a = 6

Answer 9

hate to interrupt guys... i know whpalmer is the senior one.. but u guys r missing something... what Brad tryingto say is
\[\log_{2} \frac{ 1 }{ 64 }\]

Answer 10

No, we're not missing that. it's easier to explain the positive log case first...

Answer 11

but thanks for making sure!

Answer 12

u can use
\[\log \frac{ a }{ b } = \log a - \log b\]

Answer 13

WOW u r awesome then !!

Answer 14

such a wise response certainly deserves a medal :-)

Answer 15

@Brad1996 do you see why the \[\log_2 64 = 6\]?

Answer 16

Log still confuses me... I mean what does it mean when you ask the log of something?

Answer 17

Okay, good question. Logs are like exponents, except with a fixed choice for the base. As you recall, with an exponential like \(a^b\) \(a\) is the base, and \(b\) is the exponent. When we decide to take logarithms, we choose 1 value for the base, that's the base of the logarithm, in this problem it happens to be 2.

Now, finding the log of \(n\) is just finding the exponent that you need to raise your base to to get \(n\):

\[\log_b a = n\]\[b^n = a\]
\[\log_2 64 = 6\]\[2^6 = 64\]\[\log_2 1 = 0\]\[2^0 = 1\]
We can't ever have the log of a negative number, or 0, because there is no exponent to which we could raise any base to give us either of those answers.

Answer 18

thanks, still a bit confusing but I'm sure I'll get it through practice :)

Answer 19

I'm going to post another question and try to work it through, could you help me through it?

Answer 20

One of the nifty things about logs is that it turns multiplication and division problems into addition and subtraction problems, and raising numbers to powers turns into multiplication.

Wait! You haven't finished here yet.

Answer 21

We did the log of 64, but we need the log of 1/64 to answer this question!

Do you remember the property of exponents that \[a^{-n} = \frac{1}{a^n}\]

Answer 22

Yeah I think so

Answer 23

so it would be -6?

Answer 24

because 2^-6 = 1/64?

Answer 25

Yep! Excellent!

Answer 26

Sweet :D

Answer 27

The other guy made an important point about logs that we could have used here, would you like me to elaborate on it? I'll still look at your other problem, of course.

Answer 28

sure

Answer 29

Logs turn multiplication and division into addition and subtraction.

For example, to compute 32*8, we could find \(\log_2 32\), \(\log_2 8\), add them, and take the anti-log \(2^n\). What are those two logs?

Answer 30

you lost me there bro lol

Answer 31

Well, just figure out the two logs for me:
\[\log_2 32 = \]\[\log_2 8 = \]

Answer 32

2^3 = 8
2^5 = 32

Answer 33

OS being a bit slow for me

Answer 34

Yeah, me too! :(

Okay, so if \(2^3=8\), what is \(\log_2 8 =\)
If \(2^5 = 32\), what is \(\log_2 32 = \)