Official Solution:If \(x\) is a positive integer, which of the following is closest to \(\frac{22^{22x} - 22^{2x} }{11^{11x} - 11^x}\)? A. \(2^{11x}\)
B. \(11^{11x}\)
C. \(22^{11x}\)
D. \(2^{22x}*11^{11x}\)
E. \(2^{22x}*11^{22x}\)
First of all, it's important to note that we are looking for an
approximate value of the given expression.
Now, considering that \(22^{22x}\) is significantly larger than \(22^{2x}\), the value of \(22^{22x}-22^{2x}\) will be very close to just \(22^{22x}\). In this context, \(22^{2x}\) becomes negligible. Similarly, \(11^{11x}-11^x\) will be almost equivalent to \(11^{11x}\)
Therefore, \(\frac{22^{22x}-22^{2x} }{11^{11x}-11^x} \approx \frac{22^{22x} }{11^{11x} } = \frac{2^{22x}*11^{22x} }{11^{11x} } = 2^{22x}*11^{11x}\).
For further verification, one can also approach this algebraically: \(\frac{22^{22x}-22^{2x} }{11^{11x}-11^x} = \frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}\). Since the -1 in both the denominator and numerator is negligible, the expression simplifies to: \(\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)} \approx \frac{22^{2x}*22^{20x} }{11^x*11^{10x} } = \frac{22^{22x} }{11^{11x} } = 2^{22x}*11^{11x}\).
Answer: D