Question

64=(1/8)-2

Answer 1

Not true. o.o

Answer 2

no no I'm trying to write it in Logarithmic form

Answer 3

Err, did you write the problem correctly? There is no carat '^' in your problem anywhere. Logarithms work with exponents.

Answer 4

Oh, now I see it:
\[64 = (\frac{1}{8})^{-2}\]

Answer 5

yes I was going to retype it

Answer 6

Check this out:
\[b = x^{a} ---> \log _{x} b = a\]

Answer 7

Logarithms give us the EXPONENT of a problem. There fore, if we only had the numbers 64 and (1/8), we could find the -2.

Answer 8

For example, I'll give you this logarithm:
\[\log _{10}100 = ?\]

Answer 9

This is saying, 10 to what power gives you the answer of 100. Do you know what the power would be?

Answer 10

10

Answer 11

Well, 10*10 = 100, how can you write it as an exponent? 10^? = 100

Answer 12

The exponent is kinda like how many 10's did you need to get to 100. ...

Answer 13

Since 10*10 = 100, we needed two 10's multiplied by each other to get to 100, right?
So, we can rewrite it like \[10^{2} = 100\]

Answer 14

oh

Answer 15

Therefore,
\[\log _{10}100 = 2\]

Answer 16

Ask yourself, how many '10s' do I need to multiply to get to 100.

Answer 17

ok

Answer 18

Maybe @AccessDenied can explain it better...I'm useless.

Answer 19

no you're not useless it's making sense sorry

Answer 20

So, do you have an idea on the main problem so far?
\(64 = \left( \dfrac{1}{8} \right)^{-2} \)

Answer 21

Comparing it with other examples may reinforce where these problems tend to go:
\( 100 = (10)^2 \qquad \to \qquad \log_{10} 100 = 2 \)
\( 64 = \left( \dfrac{1}{8} \right) ^{-2} \quad \ \to \)