# Double variable equation- how to solve?

1. Mar 24, 2010

### BR24

ran into this equation and wondering how to solve it. currently doing pre calculus in high school and don't think i have learned this yet but i was curious as to how to solve it and the answer. 36^x=9(32^y). any help and if you have advice on how to do this could you list the answer below so i can verify i am doing it right?

2. Mar 24, 2010

### mathman

You can get a linear relationship between x and y by taking logs on both sides.
x*log(36)=log(9)+y*log(32) (* means multiply)

3. Mar 24, 2010

### BR24

i follow how you managed to get that far, although i forgot to mention but i am supposed to find the values of x and y.

4. Mar 24, 2010

### rasmhop

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. Mar 24, 2010

### BR24

so, if i wanted to state the values of x and y, would this look correct.
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. Mar 24, 2010

### rasmhop

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:
$$x=\frac{\log(9)}{\log(36)}+\frac{\log(32)}{\log(36)}y$$
which is a linear relationship and as simple as you get.

7. Mar 25, 2010

### Anonymous217

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