Question

im given a table of x and y points and i need to find the logarithmic equation that relates the two. how do i even start?

Answer 1

Okay, well you just need to find the base then. What is the base of the logarithm? So basically if b is your base, x is your first column and y is your second, then b^y=x.

Answer 2

So if you raise both columns to the (1/y) power, then you get b ^ 1 = x ^(1/y)

Answer 3

General form for a logarithmic equation:

\[y = log_bx\]

So the first interesting point is where y = 1. Because when y = 1, what will your x be?

Answer 4

What are the first few figures of x and y that u have?

Answer 5

x=1,x=2,x=3,x=4,x=5,x=6 when y=1, y=1.189, y=1.316, y=1.414, y=1.495, y=1.565

Answer 6

Everybody seems to have disappeared.

The difficulty here is that there is not a logarithmic relation between x = 1 and y = 1
and I am trying to think what else the question could mean.

Answer 7

well thats a formula in my notes that says lny=mlnx

Answer 8

ln y = m ln x ?

Answer 9

yes

Answer 10

It should be
\[y = m(ln x)\]

And you're trying to find m.

So if you have any x,y pair you can solve for m.

Answer 11

thats what i did. im not really sure how to check it though

Answer 12

So you thought to ask us:-)

Answer 13

well when i first posted it i hadnt found that formula

Answer 14

Actually I'm not sure how that can be right, cause we still have the whole problem with (1,1). If x is 1, then we have:

\[1 = m(ln\ 1)\]

Which has no solutions for m since the ln(1) = 0

Answer 15

Do you have the actual question handy?

Answer 16

find the logarthmic equation that relates y and x.
then theres the table of points i gave earlyer

Answer 17

x=1,x=2,x=3,x=4,x=5,x=6 when y=1, y=1.189, y=1.316, y=1.414, y=1.495, y=1.565

Answer 18

I wonder could it be log something +1....

Answer 19

how can we?
shouldn't most of those y's be irrational?

Answer 20

I think they are just approximations...

Answer 21

i dont like approximations when we are trying to find an equation to relate x to y

Answer 22

Welcome to the real world:-)

Answer 23

we can only find an approximate equation theen

Answer 24

If it's ln (or log) +1 that gets us out of the 0 problem..

Answer 25

The numbers roughly behave like a log o some sort...

Answer 26

y=x^(1/4)

Answer 27

thats the approximate relation

Answer 28

i remember 2^(1/2)=1.41 blah blah
i seen that number for x=4
so i was like 4^(1/4)=1.41 blah blah
does this pattern work for the rest and then i was like oh yeah it does!

Answer 29

why we were talking about log?

Answer 30

That's the question, find the logarithmic relation, not just the relation.

And the (1,1) stumped us.

It looks a bit like lnx shuftied up a bit.

Answer 31

i need to find log equations

Answer 32

so tthe relations is y=x^1/4?

Answer 33

yes

Answer 34

so whats the log equation ?

Answer 35

no log

Answer 36

Maybe it's just log +1.

Answer 37

So I suppose then that:
\[ln y = \frac{1}{4}lnx\]

Which is the form he was asking for originally.. Just seems like a silly way to write that relation.

Answer 38

well yeah you can take log of both sides of y=x^(1/4)

Answer 39

Then we have 1 , 1.3, 1.47

Answer 40

But then it goes up too fast....

Answer 41

what are you talking about?

Answer 42

If you take log (ordinary log not ln) +1 as the relation..

Answer 43

y=x^{1/4}
1^{1/4}=1
2^{1/4}=1.189207115
3^{1/4}=1.316074013
4^{1/4}=1.414213562
5^{1/4}=1.495349781
6^{1/4}=1.56508458
honestly i have no clue why we are talking about log?

Answer 44

The question asks for the logarithmic relation between the 2 lists.

Answer 45

So the first problem is (1,1) as was discussed at the beginning...

Answer 46

then
\[\ln(y)=\frac{1}{4}\ln(x)\]

Answer 47

yeah but whats the point in saying that

Answer 48

none whatsoever !

Answer 49

I guess...

Answer 50

thats what the question asks for

Answer 51

why would we make the equation more ugly just because it ask for a log form

Answer 52

some math teacher asked the question and i guess that is what they want. just a guess though because i lost most of my mind reading powers when they got shot off in the war

Answer 53

I suspect it's a very practical class...

Answer 54

satellite am i suppose to click on that virus?

Answer 55

I have figured out when he puts that up:-)

Answer 56

im not sure who that dude is

Answer 57

"I hold in my hand the last envelope"

Answer 58

Yes, I think we have dealt with this question, after much ado, as they say.

Answer 59

Happy, Swiker?

Answer 60

yes thankyou