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Re: Variable in Exponent DS problem [#permalink]
07 Dec 2010, 17:50

Expert's post

tonebeeze wrote:

I would appreciate it if someone could walk me through this problem. Thanks

Is \(5^k\) less than 1000?

1. \(5^k+1 > 3000\) 2. \(5^k-1 = 5^k -500\)

I am assuming the question is:

Is \(5^k\) less than 1000?

1. \(5^{k+1} > 3000\) 2. \(5^{k-1} = 5^k -500\)

\(5^4 = 625\) and \(5^5 = 3125\) (even if you do not know this, it is fine. You don't need to calculate. Just observe that 625*5 will be greater than 3000)

Statement 1: \(5^{k+1} > 3000\) This means k + 1 is greater than 4 so k is greater than 3 (It doesnt mean that k + 1 is at least 5 because the question doesn't say that k is an integer. k + 1 could be 4.999 making k = 3.999) Since k can take values less than 4 and more than 4, 5^k could be less than 1000 or more than 1000. Not sufficient.

Statement 2: \(5^{k-1} = 5^k -500\) Re-arrange: \(500 = 5^k -5^{k-1}\) \(5^3 *4 = 5^{k-1}(5 - 1)\) Hence k - 1 = 3 and k = 4 So \(5^k\) = 625 which is less than 1000. Answer is 'Yes'. Sufficient.

Re: Variable in Exponent DS problem [#permalink]
03 Jan 2011, 07:29

Expert's post

m990540 wrote:

Can you go over how you went from 500 = 5^k -5^{k-1} to 5^3 *4 = 5^{k-1}(5 - 1)?

Thanks so much in advance!

Left hand side: \(500 = 5*100 = 5*25*4 = 5^3*4\) (Since your concern is the power of 5, separate 5s from the rest)

Right hand side: \(5^k -5^{k-1} = 5^{k - 1} ( 5 - 1)\)(Take \(5^{k - 1}\) common. e.g. if you have \(5^4 - 5^3\), you can take \(5^3\) common and you will be left with (5 - 1)) _________________

Re: Variable in Exponent DS problem [#permalink]
22 May 2014, 22:11

VeritasPrepKarishma wrote:

tonebeeze wrote:

I would appreciate it if someone could walk me through this problem. Thanks

Is \(5^k\) less than 1000?

1. \(5^k+1 > 3000\) 2. \(5^k-1 = 5^k -500\)

I am assuming the question is:

Is \(5^k\) less than 1000?

1. \(5^{k+1} > 3000\) 2. \(5^{k-1} = 5^k -500\)

\(5^4 = 625\) and \(5^5 = 3125\) (even if you do not know this, it is fine. You don't need to calculate. Just observe that 625*5 will be greater than 3000)

Statement 1: \(5^{k+1} > 3000\) This means k + 1 is greater than 4 so k is greater than 3 (It doesnt mean that k + 1 is at least 5 because the question doesn't say that k is an integer. k + 1 could be 4.999 making k = 3.999) Since k can take values less than 4 and more than 4, 5^k could be less than 1000 or more than 1000. Not sufficient.

Statement 2: \(5^{k-1} = 5^k -500\) Re-arrange: \(500 = 5^k -5^{k-1}\) \(5^3 *4 = 5^{k-1}(5 - 1)\) Hence k - 1 = 3 and k = 4 So \(5^k\) = 625 which is less than 1000. Answer is 'Yes'. Sufficient.

Answer (B)

I can't get why are we talking about decimals in the first statement. I don't know if I'm doing something wrong here...

I did it like this \(5^{k+1} > 3000\) \(k+1 \geq 5\) \(k \geq 4\)

\(k= 4\) then \(5^4 = 625\) \(k=5\) then \(5^5 = 3000\) (approx)

Re: Variable in Exponent DS problem [#permalink]
23 May 2014, 01:12

Expert's post

b2bt wrote:

VeritasPrepKarishma wrote:

tonebeeze wrote:

I would appreciate it if someone could walk me through this problem. Thanks

Is \(5^k\) less than 1000?

1. \(5^k+1 > 3000\) 2. \(5^k-1 = 5^k -500\)

I am assuming the question is:

Is \(5^k\) less than 1000?

1. \(5^{k+1} > 3000\) 2. \(5^{k-1} = 5^k -500\)

\(5^4 = 625\) and \(5^5 = 3125\) (even if you do not know this, it is fine. You don't need to calculate. Just observe that 625*5 will be greater than 3000)

Statement 1: \(5^{k+1} > 3000\) This means k + 1 is greater than 4 so k is greater than 3 (It doesnt mean that k + 1 is at least 5 because the question doesn't say that k is an integer. k + 1 could be 4.999 making k = 3.999) Since k can take values less than 4 and more than 4, 5^k could be less than 1000 or more than 1000. Not sufficient.

Statement 2: \(5^{k-1} = 5^k -500\) Re-arrange: \(500 = 5^k -5^{k-1}\) \(5^3 *4 = 5^{k-1}(5 - 1)\) Hence k - 1 = 3 and k = 4 So \(5^k\) = 625 which is less than 1000. Answer is 'Yes'. Sufficient.

Answer (B)

I can't get why are we talking about decimals in the first statement. I don't know if I'm doing something wrong here...

I did it like this \(5^{k+1} > 3000\) \(k+1 \geq 5\) \(k \geq 4\)

\(k= 4\) then \(5^4 = 625\) \(k=5\) then \(5^5 = 3000\) (approx)

\(5^{k+1} > 3000\) --> \(5*5^{k} > 3000\) --> \(5^{k} > 600\). Now, we are NOT told that k is an integer, thus we cannot say that \(k\geq{4}\). For example, \(5^{3.99}\approx{615}\), thus k could be 3.99.

Is 5^k less than 1,000?

Is \(5^k<1,000\)?

(1) 5^(k+1) > 3,000 --> \(5^k>600\) --> if \(k=4\) then the answer is YES: since \(600<(5^4=625)<1,000\) but if \(k=10\), for example, then the answer is NO. Not sufficient.

(2) 5^(k-1) = (5^k) - 500 --> we can solve for k and get the single numerical value of it, hence this statement is sufficient. Just to illustrate: \(5^k-5^{k-1}=500\) --> factor out \(5^{k-1}\): \(5^{k-1}(5-1)=500\) --> \(5^{k-1}=125\) --> \(k-1=3\) --> \(k=4\). Sufficient.

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