Question

Simplify: log (1/2) 8

Answer 1

(1/2)^x = 8
x = -3
log(1/2) 8 = -3

Answer 2

does "simplify" mean "evaluate?

Answer 3

Yes I think so

Answer 4

evaluate means find a value

Answer 5

Then no, it's means to simplify it as far as it can go

Answer 6

if so your job is to answer this question; one half to what power is 8, or "what would i raise one half to , to get the number 8?"

Answer 7

in other word solve for y
\[(\frac{1}{2})^y=8\]

Answer 8

Then no, it's means to simplify it as far as it can go

Answer 9

Ohkay, how do I do that??

Answer 10

and the answer is -3 pretty much by inspection

Answer 11

2 cubed is 8, and you need to flip it

Answer 12

you have to take the reciprocal and then cube. you pretty much have to figure it out from scratch, there is no formula.

Answer 13

but if you have a calculator i can show you how to cheat

Answer 14

Then no, it's means to simplify it as far as it can go

Answer 15

Ohh I see. That makes sense

Answer 16

it is easy to do if you are allowed to use a calculator

Answer 17

You should show me how to cheat :) lol

Answer 18

ok your calculator already has a log on it. it just says "log"

Answer 19

\[\large \log_{\frac{1}{2}}\left(8\right)\]



\[\large \frac{\log_{10}\left(8\right)}{\log_{10}\left(\frac{1}{2}\right)}\]



\[\large \frac{\log_{10}\left(2^{3}\right)}{\log_{10}\left(2^{-1}\right)}\]



\[\large \frac{3\log_{10}\left(2\right)}{-1\log_{10}\left(2\right)}\]



\[\large \frac{3}{-1}\]



\[\large -3\]



So \[\large \log_{\frac{1}{2}}\left(8\right)=-3\]

Answer 20

yeah i know how to use log i just don't know how to plug in th sub (1/2) thing

Answer 21

type in \[\log(8)\div\log(\frac{1}{2})\] and you will get your answer

Answer 22

You cannot. You have to change the base

Answer 23

Base! That was the word I was looking for. Does that trick work with every problem or just ones that it happens to work on like this one?

Answer 24

you can find the log of any number for any base using this
\[\log_b(x)=\log(x)\div\log(b)\] where log means the log on your calculator

Answer 25

Then no, it's means to simplify it as far as it can go

Answer 26

Okay cool. I think I remember that formula

Answer 27

if i want
\[\log_7(95)\] i just type
\[\log(95)\div\log(7)\] and i am done

Answer 28

\[log_b (a) = k \implies b^k = a\]
Taking the log (base 10) of both sides we have:
\[log(b^k) = log(a)\]\[\implies k\cdot log(b) = log(a)\]\[\implies k = \frac{log(a)}{log(b)}\]

But we started with the premise that \(log_b(a) = k\). Therefore:

\[log_b(a) = \frac{log(a)}{log(b)}\]

Answer 29

so rather than spending all my time figuring out this to the that is the other, i can just use the "change of base" formula, which is what polpak wrote

Answer 30

Yep. Though I cannot usually remember the formula, so I have to derive it every time ;p

Answer 31

i don't believe you. you want to solve
\[b^x=A \]for x i be you go right to
\[x=\frac{\log(A)}{\log(b)}\] without thinking

Answer 32

Nope. I'd have to plod along (I'd probably use ln too)

\[\implies ln(b^x) = ln(A) \implies x = \frac{ln(A)}{ln(B)}\]

Answer 33

err B=b

But even then this isn't the situation we're looking at. We're talking about \[x = log_b45\]

Answer 34

And yeah, I'm gonna have to go to the derivation for that also

Answer 35

Wonderful, I wrote that down in my notes so I don't forget.