GreenlightTestPrep wrote:
\(x\) is a positive integer, and \(x^x = \sqrt{x^{16}}\)
Quantity A |
Quantity B |
\(x\) |
\(4\) |
Rewrite the equation as follows: \(x^x = (x^{16})^{\frac{1}{2}}\)
- Taking the square root of a value is equivalent to raising that value to the power of \(\frac{1}{2}\)Simplify the right side to get: \(x^x = x^8\)
Note: I created this question to remind students that, when it comes to equations with variables in the exponents, there are three important provisos we must consider before we can conclude that two exponents are equal.
That is, if \(x^a = x^b\), then we can conclude that \(a = b\) AS LONG AS \(x \neq 0\), \(x \neq 1\), and \(x \neq -1\).
For example, if we know that \(0^x = 0^3\), we can't then conclude that \(x = 3\), since there are infinitely many values of \(x\) that satisfy the equation. So, if \(x\) does NOT equal -1, 0 or 1, then we can conclude that \(x = 8\), in which case
Quantity A is greater than Quantity BNow let's test the forbidden numbers (i.e., -1, 0 and 1).
Since we're told
x is positive, we need only test whether \(x = 1\) satisfies the given equation.
Plug \(x = 1\) into the equation \(x^x = x^8\) to get: \(1^1 = 1^8\)....IT WORKS!!
So, \(x = 1\) is another solution to the equation, in which case
Quantity B is greater than Quantity A Answer: D