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If \(r\) is a non-zero number what is the value of \(r\)?
(1) \(r^r = r\)
(2) \((\frac{1}{r})^r = r\)
(1) \(r^r = r\)
(2) \((\frac{1}{r})^r = r\)
16 Sep 2014, 00:25
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Official Solution:
If \(r\) is a non-zero number what is the value of \(r\)?
(1) \(r^r = r\)
By dividing both sides by \(r\), the equation \(r^r=r\) can be simplified to \(r^{r-1}=1\) . This leads to three possible cases:
Case 1: The base is 1, since \(1^m=1\) for all \(m\).
\(r=1\). In this case, \(r^{r-1} = 1^0 = 1\), which is true.
Case 2: The exponent is 0, since \(n^0=1\) for any nonzero \(n\).
\(r-1=0\), which implies \(r=1\). This also leads to the same solution as in Case 1.
Case 3: The base is -1 and the exponent is even, since \((-1)^k=1\) for any even \(k\).
If \(r=-1\), then \(r^{r-1}=1\) will be true since \(r-1\) will be even. Therefore, \(r=-1\) is also a possible solution.
Therefore, the solutions to the equation \(r^r=r\) are \(r=1\) and \(r=-1\).
Not sufficient.
(2) \((\frac{1}{r})^r = r\)
The above ijmplies that \(\frac{1}{r^r} = r\).
Cross-multiplying gives \(r^{r+1}=1\). This leads to the same three possible cases:
Case 1: The base is 1, since \(1^m=1\) for all \(m\).
\(r=1\). In this case, \(r^{r+1} = 1^2 = 1\), which is true.
Case 2: The exponent is 0, since \(n^0=1\) for any nonzero \(n\).
\(r+1=0\), which implies \(r=-1\). This leads to \((-1)^{0}=1\), which is true.
Case 3: The base is -1 and the exponent is even, since \((-1)^k=1\) for any even \(k\).
\(r=-1\), is the same solution as in Case 2.
Therefore, the solutions to the equation \((\frac{1}{r})^r = r\) are \(r=1\) and \(r=-1\).
Not sufficient.
(1)+(2) Each statement gives the same two solutions, \(r=1\) or \(r=-1\). Therefore, even taken together the statements are not sufficient.
Answer: E
If \(r\) is a non-zero number what is the value of \(r\)?
(1) \(r^r = r\)
By dividing both sides by \(r\), the equation \(r^r=r\) can be simplified to \(r^{r-1}=1\) . This leads to three possible cases:
Case 1: The base is 1, since \(1^m=1\) for all \(m\).
\(r=1\). In this case, \(r^{r-1} = 1^0 = 1\), which is true.
Case 2: The exponent is 0, since \(n^0=1\) for any nonzero \(n\).
\(r-1=0\), which implies \(r=1\). This also leads to the same solution as in Case 1.
Case 3: The base is -1 and the exponent is even, since \((-1)^k=1\) for any even \(k\).
If \(r=-1\), then \(r^{r-1}=1\) will be true since \(r-1\) will be even. Therefore, \(r=-1\) is also a possible solution.
Therefore, the solutions to the equation \(r^r=r\) are \(r=1\) and \(r=-1\).
Not sufficient.
(2) \((\frac{1}{r})^r = r\)
The above ijmplies that \(\frac{1}{r^r} = r\).
Cross-multiplying gives \(r^{r+1}=1\). This leads to the same three possible cases:
Case 1: The base is 1, since \(1^m=1\) for all \(m\).
\(r=1\). In this case, \(r^{r+1} = 1^2 = 1\), which is true.
Case 2: The exponent is 0, since \(n^0=1\) for any nonzero \(n\).
\(r+1=0\), which implies \(r=-1\). This leads to \((-1)^{0}=1\), which is true.
Case 3: The base is -1 and the exponent is even, since \((-1)^k=1\) for any even \(k\).
\(r=-1\), is the same solution as in Case 2.
Therefore, the solutions to the equation \((\frac{1}{r})^r = r\) are \(r=1\) and \(r=-1\).
Not sufficient.
(1)+(2) Each statement gives the same two solutions, \(r=1\) or \(r=-1\). Therefore, even taken together the statements are not sufficient.
Answer: E
General Discussion
