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m and n are integers. [#permalink]
07 Dec 2017, 03:16

Expert's post

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A

B

C

D

E

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m and n are integers.

Quantity A

Quantity B

(\sqrt{10^{2m}})(\sqrt{10^{2n}})

10^{mn}

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.

Re: m and n are integers. [#permalink]
08 Dec 2017, 05:13

1

This post received KUDOS

Hello,

for this question it is best to try out different values.

For instance.

If m and n are both zero, both expressions will be equal to 1 (since anything exponent zero is one) => Answer C If m and n are both 3, Qty A becomes 10^6 and Qty B becomes 10^9 => Answer B,

since the square root and exponent 2 cancel each other out, Qty A becomes 10^m * 10^n => 10^(m+n)= 10^(3+3)<10^(3*3)

Re: m and n are integers. [#permalink]
08 Dec 2017, 10:59

2

This post received KUDOS

Expert's post

Carcass wrote:

m and n are integers.

Quantity A

Quantity B

(\sqrt{10^{2m}})(\sqrt{10^{2n}})

10^{mn}

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.

RULE: √x = x^(1/2)

So, √[10^(2m)] = [10^(2m)]^(1/2) = 10^m [once we apply the Power of a Power rule]

Likewise, √[10^(2n)] = [10^(2n)]^(1/2) = 10^n [once we apply the Power of a Power rule]

We get:

QUANTITY A: (10^m)(10^n) QUANTITY B: 10^(mn)

Simplify to get: QUANTITY A: 10^(m+n) QUANTITY B: 10^(mn)

Let's test some values of m and n

m = 0 and n = 0 We get: QUANTITY A: 10^(0+0) = 10^0 = 1 QUANTITY B: 10^[(0)(0)] = 10^0 = 1 In this case the two quantities are equal

m = 1 and n = 1 We get: QUANTITY A: 10^(1+1) = 10^2 = 100 QUANTITY B: 10^[(1)(1)] = 10^1 = 10 In this case Quantity is greater

Answer: D

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