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What is the value of m m in the equation below?

Expert replies
by sgraves » Wed Jun 01, 2016 5:08 pm
Image


I'm having a hard time figuring out how to approach the first step in answering exponent questions, finding common bases in the above problem.
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Source: — Problem Solving |

by MartyMurray » Thu Jun 02, 2016 3:24 am
On the left side of the equation in the denominators you have a prime, 5, and a square of a prime, 4.

On the right side you have 2, a prime, and 10, a non prime.

Try factoring the denominators into primes and seeing where that takes you.
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by GMATGuruNY » Thu Jun 02, 2016 3:38 am
Every prime factor on the right side must also appear on the left side.
On the right side we have 1/10�� = 1/2�� * 1/5��.
Thus, the left side must also include 1/5��, implying that m=47.

The correct answer is D.
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by 800_or_bust » Thu Jun 02, 2016 5:59 am
sgraves wrote:Image


I'm having a hard time figuring out how to approach the first step in answering exponent questions, finding common bases in the above problem.
(1/5)^m x (1/4)^24 = 1/(2(10)^47)

(1/4)^24 can be rewritten in its equivalent form of (1/2)^48.

Now we have (1/5)^m x (1/2)^48 = 1/(2(10)^47).

So on the right side of the equation we have 47 factors of 10 in the denominator and one leftover 2.

Hence m must equal 47.

(1/5)^47 x (1/2)^48 would yield 47 factors of 10 and one leftover 2.
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by Matt@VeritasPrep » Tue Jun 07, 2016 10:53 pm
You can also crossmultiply. That gives

2 * 10�� = 5� * 4²�

which is much easier for most students to solve.
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by danielle07 » Sat Sep 02, 2017 1:48 am
It is very informative and easy to solve when we use simplified solution like this

cross multiplication

2 * 10�� = 5� * 4²�
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sgraves wrote: ↑
Wed Jun 01, 2016 5:08 pm
Image


I'm having a hard time figuring out how to approach the first step in answering exponent questions, finding common bases in the above problem.
Simplifying the equation, we have:

1/(5^m) * 1/(4^24) = 1/[2(10)^47]

1/(5^m * 2^48) = 1/[2(2^47 * 5^47)]

5^m * 2^48 = 2^48 * 5^47

5^m = 5^47

m = 47

Answer: D

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