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09 May 2024, 00:42
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If \(27^{x^2}=\frac{3}{3^{3y^2+1}}\), what is the value of \(x^2+y^3\)?
A. -3
B. -1
C. 0
D. 1
E. 3
A. -3
B. -1
C. 0
D. 1
E. 3
09 May 2024, 00:42
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Official Solution:
If \(27^{x^2}=\frac{3}{3^{3y^2+1}}\), what is the value of \(x^2+y^3\)?
A. -3
B. -1
C. 0
D. 1
E. 3
\(27^{x^2}=\frac{3}{3^{3y^2+1}}\).
\(3^{3x^2}=\frac{3}{3^{3y^2}*3}\);
\(3^{3x^2}*3^{3y^2}=1\);
\(3^{3x^2+3y^2}=1\);
Since the power of 3 must be zero in order for the above equation to hold true then \(3x^2+3y^2=0\), so \(x^2+y^2=0\). The sum of two non-negative values is zero means that both \(x\) and \(y\) must be zero therefore \(x=y=0\) and \(x^2+y^3=0\).
Answer: C
If \(27^{x^2}=\frac{3}{3^{3y^2+1}}\), what is the value of \(x^2+y^3\)?
A. -3
B. -1
C. 0
D. 1
E. 3
\(27^{x^2}=\frac{3}{3^{3y^2+1}}\).
\(3^{3x^2}=\frac{3}{3^{3y^2}*3}\);
\(3^{3x^2}*3^{3y^2}=1\);
\(3^{3x^2+3y^2}=1\);
Since the power of 3 must be zero in order for the above equation to hold true then \(3x^2+3y^2=0\), so \(x^2+y^2=0\). The sum of two non-negative values is zero means that both \(x\) and \(y\) must be zero therefore \(x=y=0\) and \(x^2+y^3=0\).
Answer: C
09 Sep 2024, 17:46
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Why is it that the power of 3 must be zero?
