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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calendarEducator since 2013
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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...
(The entire section contains 2 answers and 564 words.)
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calendarEducator since 2010
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We'll work on the 1st equation. We notice that the bases of the multiplied factors are matching, so we'll add the exponents:
4^(x/y)*4^(y/x) = 4^(x/y + y/x) = 4^[(x^2 + y^2)/xy]
But 4 = 2^2 and we'll raise to the exponent [(x^2 + y^2)/xy], both sides:
4^[(x^2 + y^2)/xy] = 2^2*[(x^2 + y^2)/xy]
32 = 2^5
The 1st equation will become:
2^2*[(x^2 + y^2)/xy] = 2^5
Since the bases are matching, we'll apply one to one property:
2*[(x^2 + y^2)/xy] = 5
We'll note x/y = z => y/x = 1/z
2(z + 1/z) = 5
2z^2 - 5z + 2 = 0
We'll determine the roots:
z1 = [5+sqrt(25 - 16)]/4
z1 = (5+3)/4
z1 = 2
z2 = 1/2
We'll put x/y = z1 <=> x/y = 2 => x = 2y
x/y = 1/2 => y = 2x
We'll re-write the 2nd equation, using the product rule of logarithms:
log3 (x-y) + log3 (x+y) = 1
log3 (x^2 - y^2) = 1
We'll take antilogarithms:
x^2 - y^2 = 3
We'll put x = 2y:
4y^2 - y^2 = 3
3y^2 = 3
y^2 = 1
y1 = 1 => x = 2
y2 = -1 => x = -2
We'll put x = y/2:
y^2/4 - y^2 = 3
y^2 - 4y^2 = 12
-3y^2 = 12
y^2 = -4
y1 = 2i => x1 = 2i/2 = i
y2 = -2i => x = -i
Since x and y values have to be real for the logarithms to exist, we'll accept only the real solutions of the equation. Furthermore, the logarithms must be positive and the only solution that satifies all these requirements is (2, 1)
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