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Which of the following is equal to (2^k)(5^k − 1)?

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Which of the following is equal to (2^k)(5^k − 1)? [#permalink] New post 11 Feb 2013, 07:50
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Which of the following is equal to 2^k*5^(k-1)?

A. 2*10^(k-1)
B. 5*10^(k-1)
C. 10^k
D. 2*10^k )
E. 10^(2k-1)
[Reveal] Spoiler: OA
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink] New post 20 Feb 2013, 20:15
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mp2469 wrote:
Bunuel wrote:
Which of the following is equal to 2^k*5^(k-1)?

A. 2*10^(k-1)
B. 5*10^(k-1)
C. 10^k
D. 2*10^k )
E. 10^(2k-1)

2^k*5^{k-1}=(2*2^{k-1})*5^{k-1}=2*10^{k-1}.

Answer: A.

I don't understand how you get 2*2^(K-1). I'm obviously missing something but can't figure it out. Can you please explain?


2^3 = 2*2*2 = 2*2^2

Similarly, 2^{10} = 2*2^9 = 2^2*2^8 etc

Hence 2^k = 2*2^{k-1} = 2^2*2^{k-2} = 2^3*2^{k-3} etc

Another Approach: Number Plugging.

Put k = 1 in 2^k*5^{k-1}. You get 2^1*5^0 = 2

When you put k = 1 in the options, only option (A) gives you 2.
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink] New post 21 Feb 2013, 02:20
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mp2469 wrote:
Bunuel wrote:
Which of the following is equal to 2^k*5^(k-1)?

A. 2*10^(k-1)
B. 5*10^(k-1)
C. 10^k
D. 2*10^k )
E. 10^(2k-1)

2^k*5^{k-1}=(2*2^{k-1})*5^{k-1}=2*10^{k-1}.

Answer: A.

I don't understand how you get 2*2^(K-1). I'm obviously missing something but can't figure it out. Can you please explain?


Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
a^n*b^n=(ab)^n

Thus, 2*2^{k-1}=2^{1+k-1}=2^k.

For more check here: math-number-theory-88376.html

Hope it helps.
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink] New post 11 Feb 2013, 07:54
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink] New post 20 Feb 2013, 17:16
Bunuel wrote:
Which of the following is equal to 2^k*5^(k-1)?

A. 2*10^(k-1)
B. 5*10^(k-1)
C. 10^k
D. 2*10^k )
E. 10^(2k-1)

2^k*5^{k-1}=(2*2^{k-1})*5^{k-1}=2*10^{k-1}.

Answer: A.

I don't understand how you get 2*2^(K-1). I'm obviously missing something but can't figure it out. Can you please explain?
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink] New post 17 Jan 2014, 04:06
Bunuel wrote:

Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
a^n*b^n=(ab)^n

Thus, 2*2^{k-1}=2^{1+k-1}=2^k.

For more check here: math-number-theory-88376.html

Hope it helps.


Hey Karishma, Hey Bunuel,

Till now, I have encountered this kind of problem several times.
Am I right to assume that these are the rules for simplifiying expontents like those in the questions:

2^k=2*2^{k-1} I can simplify from k to k-1.
2^{k+1}=2*2^k. I can simplify from k+1 to k

BUT I CAN'T simplify the first equation "backwards" meaning that if I see a exponent like 5^{k-1} I have to see directly that I have to get all other exponents to k-1??

I hope you get my question :D Thanks for your help

Greetings!
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink] New post 20 Jan 2014, 01:48
Expert's post
unceldolan wrote:
Bunuel wrote:

Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
a^n*b^n=(ab)^n

Thus, 2*2^{k-1}=2^{1+k-1}=2^k.

For more check here: math-number-theory-88376.html

Hope it helps.


Hey Karishma, Hey Bunuel,

Till now, I have encountered this kind of problem several times.
Am I right to assume that these are the rules for simplifiying expontents like those in the questions:

2^k=2*2^{k-1} I can simplify from k to k-1.
2^{k+1}=2*2^k. I can simplify from k+1 to k

BUT I CAN'T simplify the first equation "backwards" meaning that if I see a exponent like 5^{k-1} I have to see directly that I have to get all other exponents to k-1??

I hope you get my question :D Thanks for your help

Greetings!


What you need to do in any question depends on that particular question.

You know that 2^k=2*2^{k-1} so you can easily get 2^k down to 2^{k-1}. Also, 2^{k-1} = 2^k/2. So whether you bring the terms down to (k-1) or (k) depends on the question. Here all options involve multiplication. Hence you will need to use 2^k=2*2^{k-1}.
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink] New post 20 Jan 2014, 02:24
unceldolan wrote:
Bunuel wrote:

Operations involving the same exponents:
Keep the exponent, multiply or divide the bases
a^n*b^n=(ab)^n

Thus, 2*2^{k-1}=2^{1+k-1}=2^k.

For more check here: math-number-theory-88376.html

Hope it helps.


Hey Karishma, Hey Bunuel,

Till now, I have encountered this kind of problem several times.
Am I right to assume that these are the rules for simplifiying expontents like those in the questions:

2^k=2*2^{k-1} I can simplify from k to k-1.
2^{k+1}=2*2^k. I can simplify from k+1 to k

BUT I CAN'T simplify the first equation "backwards" meaning that if I see a exponent like 5^{k-1} I have to see directly that I have to get all other exponents to k-1??

I hope you get my question :D Thanks for your help

Greetings!


No, you could also change 5^{k-1} to \frac{5^{k}}{5}
It is a bit more complicated but may help to understand.

In this case, you would get
2^{k}*5^{k-1} = \frac{2^{k} * 5^{k}}{5} = \frac{10^{k}}{5} = \frac{10*10^{k-1}}{5} = 2*10^{k-1}
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Re: Which of the following is equal to (2^k)(5^k − 1)?   [#permalink] 20 Jan 2014, 02:24
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