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Re: a>1, b>5 [#permalink]
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a>1, b>5 [#permalink]
1
Caldas wrote:
\(a>1\), \(b>5\)
Quantity A
Quantity B
\((5b)^a\)
\((b^2)^a\)



Given:
QUANTITY A: \((5b)^a\)
QUANTITY B: \((b^2)^a\)

Key property #1: \((xy)^n = (x^n)(y^n)\)

Key property #2: \((x^a)^b = x^{ab}\)

Apply property #1 to Quantity A and property #2 to Quantity B to get:
QUANTITY A: \((5^a)(b^a)\)
QUANTITY B: \(b^{2a}\)

If we recognize that \((b^a)(b^a) = b^{2a}\), we can rewrite Quantity B as follows:
QUANTITY A: \((5^a)(b^a)\)
QUANTITY B: \((b^a)(b^a)\)

Since we're told \(b\) is POSITIVE, we know that \(b^a\) is positive, which means we can safely divide both quantities by \(b^a\) to get:
QUANTITY A: \(5^a\)
QUANTITY B: \(b^a\)

If \(b>5\) and \(a>1\), we can be certain that \(b^a > 5^a\)

Answer: B

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