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x^3>x^5
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22 Jan 2024, 00:17
22 Jan 2024, 00:17
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\(x^3>x^5\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Quantity A |
Quantity B |
x |
0 |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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Free Materials for the GRE General Exam - Where to get it!!
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Re: x^3>x^5
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22 Jan 2024, 13:01
22 Jan 2024, 13:01
1
This is testing knowledge of negative numbers and fractions as bases of exponents.
Let x = -2 < 0.
\(x^5 = (-2)^5 = -32\)
\(x^3 = (-2)^3 = -8\)
\(-8 > -32\), which satisfies \(x^3>x^5\).
Let x = 1/2 > 0.
\((\frac{1}{2})^5 = \frac{1}{32}\)
\((\frac{1}{2})^3 = \frac{1}{8}\)
\(\frac{1}{8} > \frac{1}{32}\), which also satisfies x^3>x^5.
Rule of thumb with tackling comparisons of exponents is to consider negatives AND fractions, on top of positives integers:
- If the base is a fraction: the greater the power, the closer to 0 it will get
- If the base is negative: the greater the odd power, the further from 0 it will get
Let x = -2 < 0.
\(x^5 = (-2)^5 = -32\)
\(x^3 = (-2)^3 = -8\)
\(-8 > -32\), which satisfies \(x^3>x^5\).
Let x = 1/2 > 0.
\((\frac{1}{2})^5 = \frac{1}{32}\)
\((\frac{1}{2})^3 = \frac{1}{8}\)
\(\frac{1}{8} > \frac{1}{32}\), which also satisfies x^3>x^5.
Rule of thumb with tackling comparisons of exponents is to consider negatives AND fractions, on top of positives integers:
- If the base is a fraction: the greater the power, the closer to 0 it will get
- If the base is negative: the greater the odd power, the further from 0 it will get
gmatclubot
Re: x^3>x^5 [#permalink]
22 Jan 2024, 13:01
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