# Algerbraic logarithm method question

1. Sep 20, 2009

### Cosmicon

First of all, greetings to the scientific community here at Physics Forums.

The following set of equations is given:
y^(x-3y) = x^2
x^(x-3y) = y^8

With the next assumption given: x-3y is unequal to (-4).

My attempt was to isolate the y variable in both of the equations so that i will be able to devide the the equations and recieve a simple x^a = b equations. This did not help, because the powers of the x's included the y, with no option to get rid of it.

I Would like your assistance with finding the shortest, most efficient, non-logarithmic way to solve this question, asap.

Last edited: Sep 20, 2009
2. Sep 20, 2009

### Hurkyl

Staff Emeritus
Why haven't you taken the logarithm of the equations? Isn't that the method you said you were asking about in the title?

3. Sep 20, 2009

### Cosmicon

You are right it is a possibility. I did not fully mention my request here in the topic.
I am searching for an additional method to logarithm, by modifying the powers in the two equations.

Just tackled with another question with great similarity:

x^(2x+y) = y^4
y^(2x+y) = x^16

I noticed the identity of the powers in the left sides of the equations, but even after immediatly deviding the two equations, I get stuck.

What would be the most efficient way to solve this with no log?

Last edited: Sep 20, 2009
4. Sep 20, 2009

### g_edgar

It is the method of logarithm, you just don't write log ...

$x^{(2x+y)^2} = y^{4(2x+y)} = x^{16\cdot 4} = x^{64}$. Therefore $(2x+y)^2 = 64$ (that's where we really did logarithms without mentioning the word), so assuming $x>0, y>0$ we have $2x+y=8$. Then $x^8 = y^4$ so $x^2=y$. Substitute in the equation: $2x+x^2=8$ then solve the quadratic equation. The only positive solution is $x=2$ so then $y=4$.

I assumed x and y are positive so that irrational exponents would be defined. But to solve for integer solutions, you could allow negative x,y since integer powers still make sense.

5. Sep 20, 2009

### Cosmicon

edgar, using which formula did you make this calculation:

$x^{(2x+y)^2} = y^{4(2x+y)} = x^{16\cdot 4} = x^{64}$

6. Sep 22, 2009

### g_edgar

Start with $x^{(2x+y)} = y^{4}$, raise both sides to power $2x+y$ , get $x^{(2x+y)^2} = y^{4(2x+y)}$.
Start with $y^{(2x+y)} = x^{16}$, raise both sides to the power $4$ , get $y^{4(2x+y)} = x^{16\cdot 4}$ .
OK?

7. Sep 23, 2009

### Cosmicon

yes, I now understand this technique. very useful and widespread in problems involving equations.

*According to this, when solving the quadratic equation, two answers apply: x=2, and x=-4.
why is x=-4 (y=16) not valid in this case?

** It is also important to mention that another solution in this form of questions:
when the two bases are equal to 1, the powers can be placed with any suggested number.
Therefore: when x=1, y=1.

Last edited: Sep 23, 2009