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Bunuel
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I think this is a high-quality question and I agree with explanation. Doesn't the x go away in 11^11x?? I'm getting 2^22x*11^11
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I think this is a high-quality question and I agree with explanation. Doesn't the x go away in 11^11x?? I'm getting 2^22x*11^11

Check below:

\(\frac{22^{22x}}{11^{11x}}=\frac{(2*11)^{22x}}{11^{11x}}=\frac{2^{22x}*11^{22x}}{11^{11x}}=2^{22x}*11^{22x-11x}=2^{22x}*11^{11x}\)

Does it make sense?
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This is a very nice question. I spent lots of time trying to figure out the solution. The explanation give is very nice.
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+1 for option D. One needs to use approximation here.
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I have edited the question and the solution by adding more details to enhance its clarity. I hope it is now easier to understand.
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I think this is a high-quality question and I agree with explanation.
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Bunuel... Wasted almost 4 mins to solve the euqnation.
Please provide other such questions for practice.
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I like the solution - it’s helpful.
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Hi Bunuel,

Could you explain the concept behind such questions as to when should we consider aa value negligible and in what kind of questions?

In the given below question the first instinct is to solve it wasting lot of precious time. So I just wanted to understand is there some way we can identify such questions.

Bunuel
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
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MegB07
Hi Bunuel,

Could you explain the concept behind such questions as to when should we consider aa value negligible and in what kind of questions?

In the given below question the first instinct is to solve it wasting lot of precious time. So I just wanted to understand is there some way we can identify such questions.



You should understand how exponentiation works and build number sense by practicing more problems of this type. When powers grow, the higher exponent terms completely dominate, and the smaller ones become irrelevant. That’s the idea you need to internalize instead of trying to grind through every subtraction. This kind of situation usually shows up in expressions with very large exponents or huge differences in orders of magnitude, and with enough practice, you’ll spot it immediately.

Similar questions to practice:


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Hope it helps.
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This is very helpful, thanks!
Bunuel


Check below:

\(\frac{22^{22x}}{11^{11x}}=\frac{(2*11)^{22x}}{11^{11x}}=\frac{2^{22x}*11^{22x}}{11^{11x}}=2^{22x}*11^{22x-11x}=2^{22x}*11^{11x}\)

Does it make sense?
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I like the solution - it’s helpful.
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I like the solution - it’s helpful.
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