Question

If \(log_x(y)=2\), what is the value of \(log_y(x)\)

Answer 1

If \(log_x(y)=2\), what is the value of \(log_y(x)\)

Answer 2

Any idea @AJ01?

Answer 3

This is what I have to help:
\(log_xy=2\)
\(x^2=y\)

Answer 4

Assume that x, a, and b are all positive. Also assume that a ≠ 1, b ≠ 1.

Change of base formula: \[\log_{a} x = \log_{b} x/\log_{b} a\

Answer 5

do the same for the other one

Answer 6

Wait correcting those

Answer 7

K

Answer 8

log_y(x)=a

Answer 9

should be y^a=x

Answer 10

Assume that x,a and b are all positive. Also assume that
\[a \neq 1\space ; \space b \neq 1\]

Change of base formula:\[\log_{a} x = \log_{b} x/\log_{b} a\

Answer 11

solve for y

Answer 12

x^2=y
y^a=x

Answer 13

y^a=srq(y)

Answer 14

a=1/2

Answer 15

I am really confused. Two different things. All I understand is \(y=x^2\). Now, how do I solve the question \(log_yx\)

Answer 16

okay....

Answer 17

log_y(x)=a is this okay

Answer 18

I subbed this in:
\(log_yx\)
Subbed in the valye for \(y\)
\[log_{x^2}x\]
Then did this:
\[log_{x^2}(x^2)^\frac{1}{2}\]\[=\frac{1}{2}\]That is the right answer. Is that the proper working though?

Answer 19

And I understand \(log_y(x)=a\)

Answer 20

okay write this as y^a=x

Answer 21

Yes

Answer 22

then you have 2 equation
x^2=y
y^a=x

Answer 23

put x from eq1 into eq2

Answer 24

Then solve from there?
THANK YOU

Answer 25

yes

Answer 26

Could I just ask one more question?

Answer 27

ok

Answer 28

What is \(log_kk\sqrt{k}\)

Answer 29

I did this:
\[log_kk\sqrt{k}\]\[=log_k\sqrt{k^2}\sqrt{k}\]\[=log_k\sqrt{k^3}\]\[=log_kk^\frac{3}{2}\]\[=\frac{3}{2}\]

Answer 30

Is that ok?

Answer 31

i am have problem reading the q

Answer 32

The question?

Answer 33

yes...

Answer 34

Just wanted to add a quck note for your previous question. I thought it was kind of a unpractical way to solve it.

I think this might be easier
\[2 = \log_x(y) = \frac{ \log(y) }{ \log(x) }\]\[\frac{ 1 }{ 2 } = \frac{ 1 }{ \frac{ \log(y) }{ \log(x) } } = \frac{ \log(x) }{ \log(y) } = \log_y(x)\]

Answer 35

\[\log _{k}(k \sqrt{k})\]

Answer 36

Yes

Answer 37

\[\log _{k}(k ^{3/2})\]

Answer 38

@Lyrae, that probably is more easier, its just will be hard for me to memorise and use that frequently in the test

Answer 39

So I was right, the answer was \(\frac{3}{2}\)

Answer 40

Here is one for you guys that you probably know how to do:
\[\frac{log_a81}{log_a3}\]

Answer 41

Stumped me until the teacher told me ;)

Answer 42

sorry i have problem with page..... i can't Equation form

Answer 43

OH

Answer 44

i can't read the Equation form

Answer 45

i will log out then i will log again...

Answer 46

lol @AJ01, you will log out to help me with my log problems ;)

Answer 47

\[\frac{ \log_a(81) }{ \log_a(3) } = \frac{ \log_a(3^4) }{ \log_a(3) } = \frac{ 4 \log_a(3) }{ \log_a(3) } = 4\]

Answer 48

Wow you are smat @Lyrae

Answer 49

smart*

Answer 50

Sorry if I am bothering you, but could you please help me in one question?

Answer 51

If it's a quck one :) Have to leave soon.

Answer 52

Yay
\[log_35-log_3x+log_32=log_310\]

Answer 53

Use the laws
\[\log_x a + \log_x b = \log_x(ab)\]\[\log_x a- \log_x b = \log_x(\frac{ a }{ b })\]

Answer 54

I got this far
\[log_3\frac{10}{x}=log_310\]

Answer 55

Not sure now

Answer 56

sorry but this what i see