Question

Help with Logarithms. MEDAL and FAN. Question below...

Answer 1

\[Log _{\frac{ 1 }{ 4 }}(2)=?\]

Answer 2

That is better!

Answer 3

\(\color{#000000}{ \displaystyle [1]~\quad \log_a(a)=1 }\)
\(\color{#000000}{ \displaystyle [2]~\quad \log_a(b^c)=c\times\log_a(b) }\)

Answer 4

If you know these properties, you should be got to go:)

Answer 5

I'm confused

Answer 6

\(\color{#000000}{ \displaystyle 2=2^1=2^{(-2) \times (-1/2)} =(2^{-2})^{1/2}=\left(\frac{1}{2^2}\right)^{(-1/2)} =\left(\frac{1}{4}\right)^{(-1/2)}. }\)

Answer 7

So, you can use the fact that,

\(\color{#000000}{ \displaystyle 2=\left( \frac{1}{4}\right)^{-1/2} }\)

to evaluate the logarithm (using equation [2] and [1])

Answer 8

@zepdrix

Answer 9

Would it be the opposite of the other equation where the numbers are switched?

Answer 10

The other equation being\[Log _{2}(\frac{ 1 }{ 4 })=-2\]

Answer 11

Hmm this one is a little tricky to explain :)
Didn't quite understand Zel's transformation of the 2?

So how do you turn a 2 into a 4?
By taking the square root of it, right?\[\large\rm 2=\sqrt4\]

Answer 12

I mean, by taking the square root of the 4*

Answer 13

But you would prefer to write your root as a rational exponent,\[\large\rm 2=\sqrt4=4^{1/2}\]

Answer 14

The furthest I understood from Zel's explanation was were 2 turned into\[\frac{ 1 }{ 2^2 }\]

Answer 15

Zep, all I did is wrote 2 in terms of (1/4) raised to some power.
This way I can simplify by taking the exponent out of the log.

Answer 16

Yes, I know what you did,
but he/she didn't :)

Answer 17

I did a mistake in step 4, a typo, the exponent should be -1/2 (not 1/2).
....

Ok

Answer 18

Ok, 4ever, do you understand why I needed to write the 2 as (1/4) raised to some power?

Answer 19

I understand why the 2 needed to be turned into (1/4), but I don't understand why it needs to be raised to a power

Answer 20

I need to write the 2 in terms of 1/4, but I
can't just replace the 2 with 1/4 (Right ?).

I derived that (1/4)^(-1/2) is the same thing as 2.
And so, I can use that to evaluate my logarithm.

Answer 21

ohh

Answer 22

A similar example.

\(\color{#000000}{ \displaystyle \log_{(1/9)}(3) =\log_{(1/9)}\left[3^{(-2)\times (-1/2)}\right] = \\[0.5em] }\)
\(\color{#000000}{ \displaystyle \log_{(1/9)}\left[\left(3^{-2}\right) ^{-1/2}\right] = \log_{(1/9)}\left[\left(\frac{1}{9}\right) ^{-1/2}\right] = }\)

I can use rule [2] to get the exponent out.

\(\color{#000000}{ \displaystyle (-1/2)\times \log_{(1/9)}\left[ \frac{1}{9}\right] = }\)

Then, I can use [1] to simplify more.

\(\color{#000000}{ \displaystyle (-1/2)\times 1= -1/2 }\)

Answer 23

Okay, so what do I do when I get to \[(-1/2)Log _{\frac{ 1 }{ 4 }}=\frac{ 1 }{ 4 }\]

Answer 24

How do I use [1] for the next step?

Answer 25

You mean,

\(\color{#000000}{ \displaystyle (-1/2)\times \log_{~1/4} (1/4) }\)

Answer 26

Then, you know that since, \(\color{#000000}{ \displaystyle \log_{a} (a) =1 }\)

Therefore, you also know that, \(\color{#000000}{ \displaystyle \log_{~1/4} (1/4) =1 }\)

This way, if you had (for example) \(\color{#000000}{ \displaystyle 55\times \log_{~1/4} (1/4) }\)

Then you would get: \(\color{#000000}{ \displaystyle 55\times \log_{~1/4} (1/4)=55\times (1) =55 }\)


So, in your case,
\(\color{#000000}{ \displaystyle (-1/2)\times \log_{~1/4} (1/4) =? }\)

Answer 27

(-1/2)

Answer 28

Yes, exactly!

Answer 29

negative one half :)

Answer 30

Thank you!

Answer 31

No problem:)