First answer when does \((5^x)^{5-3x} =1\)?
\((5^x)^{5-3x} = 5^{x(5-3x)}\) --> \(5^{x(5-3x)} = 5^{5x-3x^2}\\
\)
The only way for \(5^{5x-3x^2} = 1\) is if the exponent \(5x-3x^2 = 0\)
\(5x-3x^2 = 0\) --> Add \(3x^2\) to both sides
\(5x = 3x^2\) --> Divide both sides by \(3x\)
\(x = 5/3\) (this is 1 solution)
Another way to solve is to factor out x from \(5x-3x^2 = 0\)
And you get \(x(5-3x) = 0\) and this is only possible is when \(x = 0\)
Therefore \(x = 0\) or \(x = \frac{5}{3}\\
\)
Statement (1) Clearly insufficient as \(x\) could be any integer including \(0\)
Statement (2) \(x*y\)
≠ \(x\)
First determine when \(x*y = x\) by re-writing the equation
\(x*y - x = 0\) --> Now factor out x
\(x(y-1) = 0\) --> this equation can only be true if \(x = 0\) or \(y = 1\)
So we know from this statement that \(x\) cannot \(= 0\) as otherwise \(x*y\) will equal \(0\) which is equal to \(x\), but if \(y = 1\) then \(x\) could be any value including \(\frac{5}{3}\) as seen above. So this Statement alone is insufficient.
Combining Statement (1) and (2)
From statement (1) we know that x must be an integer and could be any integer including \(0\), and from statement (2) we know that integer cannot be \(0\), and since we know the only way that \((5^x)^{5-3x} =1\) is when \(x = 0\) or \(x = \frac{5}{3}\), then \((5^x)^{5-3x} ≠ 1\), thus giving us a definitive NO Answer, therefore combining both statements is sufficient to answer the question.
Answer: C.sk1999
Hi, can somebody share a detailed solution for this one?
GMATinsight
Is \(({5^x})^{5-3x} = 1\) ?
1) x is an Integer
2) x*y ≠ x