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BR24

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- Thread starter BR24
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- #1

BR24

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- #2

mathman

Science Advisor

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x*log(36)=log(9)+y*log(32) (* means multiply)

- #3

BR24

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- #4

rasmhop

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There is not really one answer, but rather infinitely many. It's like asking for THE solution to x+y=10. I could choose x=1 and y=9, or x=2 and y=8 or many others. Similarly in your case for any real number which you want x to be you can find a value of y such that your equation is satisfied (use the approach given by mathman). The same can be done if you have a specific value for y. For instance if I want y=1, then I let x=(log(9)+log(32))/log(36) and have one answer. One the other hand I might want y=2 and then I let x=(log(9)+2log(32))/log(36) and this is another valid answer.

Normally a good rule of thumb is that if you have n unknowns, then you need n equations to solve for the unknowns (this doesn't always hold true, but it's true in a lot of simple cases).

- #5

BR24

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x belongs to real numbers.

y belongs to real numbers.

are there any restrictions i would encounter in this, or would the above description be fitting. x and y do not have to be > 0 or anything along those lines right?

is there any common pattern between x and y? just as an example, something along the lines of x= 2y or something along those lines? curious as to how defined an answer i can get.

- #6

rasmhop

- 430

- 3

x belongs to real numbers.

y belongs to real numbers.

are there any restrictions i would encounter in this, or would the above description be fitting. x and y do not have to be > 0 or anything along those lines right?

is there any common pattern between x and y? just as an example, something along the lines of x= 2y or something along those lines? curious as to how defined an answer i can get.

It's easy to see that not any pair (x,y) of real numbers will work as you can just try x=y=0 which would give 1=9 which is false. The general relation is the one given by mathman. Both sides of the equation are positive so you can take the logarithm to get mathman's equation. In the same way any two numbers satisfying mathman's equation will satisfy your equation because the logarithm is injective (i.e. log(a)=log(b) if and only if a=b). x and y do not have to be positive.

Mathman provided you with the relation:

[tex]x=\frac{\log(9)}{\log(36)}+\frac{\log(32)}{\log(36)}y[/tex]

which is a linear relationship and as simple as you get.

- #7

Anonymous217

- 355

- 2

You could of course change the base of the logs and such, but the equation is simplified enough.

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