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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]

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New post 11 Feb 2013, 08:50
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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)
Spoiler: OA
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink]

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New post 11 Feb 2013, 08:54
1
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.
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink]

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New post 20 Feb 2013, 18:16
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]

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New post 20 Feb 2013, 21:15
2
1
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]

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New post 21 Feb 2013, 03:20
1
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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Collection of Questions:
PS: 1. Tough and Tricky questions; 2. Hard questions; 3. Hard questions part 2; 4. Standard deviation; 5. Tough Problem Solving Questions With Solutions; 6. Probability and Combinations Questions With Solutions; 7 Tough and tricky exponents and roots questions; 8 12 Easy Pieces (or not?); 9 Bakers' Dozen; 10 Algebra set. ,11 Mixed Questions, 12 Fresh Meat

DS: 1. DS tough questions; 2. DS tough questions part 2; 3. DS tough questions part 3; 4. DS Standard deviation; 5. Inequalities; 6. 700+ GMAT Data Sufficiency Questions With Explanations; 7 Tough and tricky exponents and roots questions; 8 The Discreet Charm of the DS; 9 Devil's Dozen!!!; 10 Number Properties set., 11 New DS set.


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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink]

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New post 17 Jan 2014, 05:06
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]

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New post 20 Jan 2014, 02:48
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]

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New post 20 Jan 2014, 03:24
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]

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New post 13 Aug 2015, 00:36
4112019 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)=10^k*5^-1
option (A) is correct
2*10^(k-1) = 10^k*5^-1
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink]

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New post 15 Aug 2015, 08:43
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.



Is the following also correct ?

2^k x 5^(k-1) = 2^(k) x 5^(k) x 5^(-1)
= 10^(k)/5

?
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink]

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New post 18 Aug 2015, 00:53
mike34170 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.



Is the following also correct ?

2^k x 5^(k-1) = 2^(k) x 5^(k) x 5^(-1)
= 10^(k)/5

?


Yes it is but it doesn't match any of the given options.
So you need to split the numerator as

\(10*10^{k - 1}/5 = 2*10^{k - 1}\)
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink]

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New post 08 Mar 2016, 07:21
4112019 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)



Nice one


=> 2^k-1 * x 2 x 5^k-1 => 10^k-1 x 2 => option A
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Re: Which of the following is equal to (2^k)(5^k − 1)? [#permalink]

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New post 11 Sep 2017, 08:43
1
4112019 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)

\(2^k*5^{(k-1)}\)

Simplifying the expression we get;

\(2^k*\frac{5^k}{5}\)

\(\frac{2^k*5^k}{5}\)

\(\frac{(2*5)^k}{5} = \frac{10^k}{5}\)

Check the options;

(A) \(2*10^{(k-1)} = 2*\frac{10^k}{10} = \frac{10^k}{5}\)

Answer (A)...
Re: Which of the following is equal to (2^k)(5^k − 1)?   [#permalink] 11 Sep 2017, 08:43
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