Algerbraic logarithm method question

[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 divide the the equations and receive 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.

 

[Hurkyl]

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

 

[Cosmicon]

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

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?

 

[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.

 

[Cosmicon]

edgar, using which formula did you make this calculation:

<br /> x^{(2x+y)^2} = y^{4(2x+y)} = x^{16\cdot 4} = x^{64}<br />

 

[g_edgar]

edgar, using which formula did you make this calculation:

<br /> x^{(2x+y)^2} = y^{4(2x+y)} = x^{16\cdot 4} = x^{64}<br />

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?

 

[Cosmicon]

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?

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.