What are x and y if 4^(x/y)*4^(y/x)=32 and log 3 (x-y)=1-log 3 (x+y) ?

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In this question, I am not clear whether the logs have the base of 3 or 3 is the part of the arguments. I will try to show both cases.

Let's consider the given system of equations:

4^(x/y)*4^(y/x) = 32

and log (3(x-y)) =...

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In this question, I am not clear whether the logs have the base of 3 or 3 is the part of the arguments. I will try to show both cases.

Let's consider the given system of equations:

4^(x/y)*4^(y/x) = 32

and log (3(x-y)) = 1 - log(3(x + y)).

First, we can use the rules of exponents to simplify the first equation. When multiplying powers of the same base, we add the exponents. Additionally, we can write 4 as 2^2 and 32 as 2^5. Then,

(2^2)^(x/y + y/x) = 2^5.

When power is taken to another power, the exponents multiply, so

2^(2(x/y + y/x)) = 2^5.

Since bases are the same, the exponents also have to be the same, so

2(x/y + y/x) = 5.

Now let's work with the second equation and use the rules of logarithms to simplify the second equation. First, rearrange the terms in the equation so that both logarithms are on the left side. (Here, I am assuming that 3 is the part of the arguments.)

log(3(x - y)) + log(3(x + y)) = 1.

When two logarithms are added, their arguments multiply. Also, 1 could be written as log of 10:

log[9(x-y)(x+y)] = log(10).

Since the logarithms of the same bases are equal, their arguments are also equal:

9(x-y)(x+y) = 10

Now we have a system of two algebraic equations:

2(x/y + y/x) = 5

and 9(x-y)(x + y) = 10.

Adding fractions in the second equation results in

2(x^2 + y^2)/xy = 5.

Assuming neither x nor y is equal to 0, we can multiply be xy and rearrange the terms:

2x^2 -5xy + 2y^2 = 0.

Factoring this by grouping results in

(x - 2y)(2x - y) = 0.

This yields x = 2y and x = y/2. Substituting the first relationship in the second equation, we get

9(2y - y)(2y + y) = 10

27y^2 = 10

y= +- sqrt(10/27) = +-sqrt(30)/9

The corresponding values of x are x = +-2sqrt(30)/9.

From the second relationship x = y/2 we get

9(y/2-y)(y/2 + y) = 10

-27y^2/4 = 10

This equation has no real solutions.

From the two pairs of solutions found above,

(2sqrt(30)/9, sqrt(30)/9) and (-2sqrt(30)/9, -sqrt(30)/9), only the first one is the possible solution. This is because the arguments x + y and x -y have to be positive, and this condition is fulfilled only by the first pair.

However, if 3 is the base of the logarithms, then the second equation will be

(x-y)(x+ y) = 3

For x = 2y, this means

(2y-y)(2y+ y) = 3

3y^2 = 3

y = +-1, with the corresponding x = +-2.

For x = y/2, we get

(y/2 - y)(y/2 + y) = 3

-3y^2/4 = 3

This equation has no real solutions.

So the possible solutions are (2, 1) and (-2, -1). Again, only the first pair works.

Depending on the problem, the answer is either

x = 2sqrt(30)/9, y = sqrt(30)/9) or

x = 2, y = 1.

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We have the equations:

4^(x/y)*4^(y/x) = 32 ...(1)

log 3 (x-y) = 1 - log 3 (x+y) ...(2)

From (1) we get

4^(x/y)*4^(y/x) = 32

=> 4^(x/y + y/x) = 2^5

=> 2^2(x/y + y/x) = 2^5

equate the exponents as the base is the same

2(x/y + y/x) = 5

=> x/y + y/x = 2.5

We can get this using x/y = 1/2 and y/x = 2

The solution sets we get are (1,2) , (2,1) (-1,-2) and (-2, -1)

But log(3)(x - y) = 1 - log(3)(x + y), so x- y and x + y should be positive. This is possible only with the solution (2, 1) as with all the other sets either x - y or x + y is negative.

The solution of the equation is only (2,1)

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