Question

9^(logbase3(20))

Answer 1

I got to (3(logbase3(20))^2

Answer 2

Yeah that works.

Answer 3

But how do I keep going here?

Answer 4

so then \(3^{log_3}=1\)

Answer 5

And all you'd be left with would be 20^2

Answer 6

?
How did that work?

Answer 7

I don't see how the original equation becomes 3^(logbase3)

Answer 8

Well, \(9 = 3^3\) right?

Answer 9

right...

Answer 10

and so now you have \[\large 3^{3 \cdot \log_3 (20)}\] which can be written as \[\large 3^{\log_3(20)^3}\] using log rules.

Answer 11

right...
but how does it convert to
3^(logbase3)=1

Answer 12

Then, you have a base that is the same as the base of the log function, and from what I remember that a base number that is raised to the log function of the same base will give you the function itself. \[\large \color{red}3^{log_{\color{red}{3}}(20)^2}\]

Answer 13

Oh I got it.

Answer 14

Think of \(\log_3 (20)^2\) as x

Therefore \(3^x\) would be the inverse of \(\log_3 (x)\)

SOO... if you have \(3^{\log_3 (x)}\) you're technically "undoing" the function

Answer 15

Do you kind of get it?

Answer 16

you can also find inverses that "undo" eachother like inverse trig functions, if you've learned them. \[\sin^{-1}(\sin(x)) = x\] But that's besides the point, just other examples of how functions undo eachother. :')

Answer 17

I guess...?
9^(logbase3(20))
(3(logbase3(20))^2
and by what you mean if 20^2 is considered x,
3^logbase3(x)=1?

Answer 18

Not quite. \[\large 9^{log_3 (20)}\]\[\large 3^{3\cdot \log_3 (20)}\]\[\large 3^{log_3 (20)^2}\]\[20^2 =?\]

Answer 19

400

Answer 20

Good job.

Answer 21

I don't see how yoou wrote 3^log3 is 1

Answer 22

You might not understand the concept of functions undoing eachother at first, but it'lljust practice and recognizing similar bases

Answer 23

well, that was conceptual. \[3^{\log_3 (20)^2} = 1 \cdot 20^2 = 20^2\]

Answer 24

oh...

Answer 25

And looking back at your explanation, you were close.

Oh, looking back at your post, I can see where you were going with it. Sorry about that.

nicklife2
I guess...?
9^(logbase3(20))
(3(logbase3(20))^2
and by what you mean if 20^2 is considered x,
3^logbase3(x)=1? < ----- you stated that x = 20^2 so I think what you meant to write was it would = 1 * 20^2

Answer 26

I don't know why I repeated the first two lines in my previous post, whoops.

Answer 27

I figured what you mean thank you.

Answer 28

Awesome :) Good luck!

Answer 29

@nicklife2

"I got to (3(logbase3(20))^2"

I think you meant to say 'Zarkon only work it out this far...can you give me the rest'

http://openstudy.com/study?source=email#/updates/54723560e4b0db93229ca7b1

Answer 30

Hmm... that's not good.