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)) = 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...

(The entire section contains 2 answers and 564 words.)

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