Question

how can i set this up equal to 0 correctly,
log𝑏(𝑣√𝑢)=log𝑏(𝑢)/𝑣

Answer 1

i know its an identity

Answer 2

just rust on how to shift things over

Answer 3

rusty*

Answer 4

Chris: Could you possibly do a screen shot and share the image with me here?

Answer 5

\[\log_{\sqrt[y]{x}} = \log_{x} /y\]

Answer 6

Seeing your image made all the difference. So much clearer!

You are beginning with \[\log_{b}\sqrt[y]{x} \]

Answer 7

right can i move the stuff over

Answer 8

from the right side

Answer 9

and want to simplify this expression. First of all, not that this can be re-written as \[\log_{b} x ^{\frac{ 1 }{ y }}\]

Answer 10

to left so it can be \[\log_{b}(\sqrt[y]{x}) (y)(\log(x))\]

Answer 11

=0

Answer 12

I don't agree that that equals 0. Why are you so sure you're supposed to end up with 0? Please note that your expression, above, is NOT an equation, nor will it become an equation. What do the instructions for this problem say?

Answer 13

oops would it be log(\[\sqrt[y]{x}\])(y)(1/log(x)) = 0

Answer 14

Hold, please. read the original problem, find the instructions and share them here.

Answer 15

they say they are the same values this values are identities the formula i gave you , so if i move everything over to one side it should give me zero on the other side right?

Answer 16

You could show that the left side equals the right side.
No, that is not right, because this problem does NOT involve addition or subtraction of terms.

Answer 17

oh i see

Answer 18

this makes sence now sorry been really rusty with math lately

Answer 19

You are beginning with:\[\log_{b}\sqrt[y]{x}\]

Answer 20

right

Answer 21

As before, you need to re-write that as \[\log_{b}x ^{\frac{ 1 }{ y }} \]

Answer 22

Do you agree with this? If not, explain why not.

Answer 23

yes i agree

Answer 24

\[\sqrt[y]{x}=x ^{\frac{ 1 }{ y }}\]

Answer 25

All right then. If you agree with\[\log_{b}x ^{\frac{ 1 }{ y }}\]

Answer 26

then we will proceed to apply this rule of logs:\[\log a^b=b*\log a\]

Answer 27

So: take \[\log_{b}x ^{\frac{ 1 }{ y }}\]

Answer 28

and rewrite it according to the rule of logs I've just given you.

Answer 29

(1/y)log(x)

Answer 30

You're not working with log x; you are working with "log to the base b of x. Please fix your response (above).

Answer 31

You want\[\log_{b}x,~NOT ~\log x \]

Answer 32

So your \[\frac{ 1 }{ y }\log x\]

Answer 33

\[(1/y)\log_{b}(x) \]

Answer 34

should be ... \[\frac{ 1 }{ y }\log_{b} x\]

Answer 35

Yes. That's the right side of your original equation. Work is done. Any questions? Note that there was NO addition or subtraction here.

Answer 36

i see i understand now so i can say if we subtract one side from the other side we should get 0

Answer 37

No, you can't. Repeat: There is NO addition, NO subtraction, here, only multiplication, division, exponentiation and rules of logs.

Answer 38

Your job is to show that the equation is true.
Forget "addition, subtraction, making it = 0." Not applicable here.

Answer 39

ok

Answer 40

Hope you are satisfied. Post another question (separately), if you like.