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What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1

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What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1  [#permalink]

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New post 14 Dec 2014, 00:41
1
6
00:00
A
B
C
D
E

Difficulty:

  95% (hard)

Question Stats:

29% (01:33) correct 71% (01:36) wrong based on 149 sessions

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What is the value of x ?

(1) 3^x X 5^y = 75
(2) 3^(x-1)(y-2) = 1
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Re: What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1  [#permalink]

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New post 14 Dec 2014, 05:15
anceer wrote:
What is the value of x ?

(1) 3^x X 5^y = 75
(2) 3^(x-1)(y-2) = 1



Here we are not told whether x is an integer or not.

st.1 3^x.5^y = 75, if x and y both are integers, then x=1 and y=2. but if they are not, then x and y can take any non-integer value. for example say x=0, then we have
5^y=75

y=2.68

st.2 3^(x-1)(y-2)=1
(x-1)(y-2)=0

if x=1 then y can have any value. if y=2 then x can have any value.

hence insufficient

combining st.1 and st.2

we have a common solution of x=1.

hence answer should be C.
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What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1  [#permalink]

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New post 15 Dec 2014, 07:59
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What is the value of \(xy\)?

Notice that we are not told that the \(x\) and \(y\) are integers.

(1) \(3^x*5^y=75\) --> if \(x\) and \(y\) are integers then as \(75=3^1*5^2\) then \(x=1\) and \(y=2\) BUT if they are not, then for any value of \(x\) there will exist some non-integer \(y\) to satisfy given expression and vise-versa (for example if \(y=1\) then \(3^x*5^y=3^x*5=75\) --> \(3^x=25\) --> \(x=some \ irrational \ #\approx{2.9}\)). Not sufficient.

(2) \(3^{(x-1)(y-2)}=1\) --> \((x-1)(y-2)=0\) --> either \(x=1\) and \(y\) is ANY number (including 2) or \(y=2\) and \(x\) is ANY number (including 1). Not sufficient.

(1)+(2) If from (2) \(x=1\) then from (1) \(3^x*5^y=3*5^y=75\) --> \(y=2\) and if from (2) \(y=2\) then from (1) \(3^x*5^y=3^x*25=75\) --> \(x=1\). Thus \(x=1\) and \(y=2\). Sufficient.

Answer: C.
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Re: What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1  [#permalink]

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New post 03 Mar 2015, 03:28
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anceer wrote:
What is the value of x ?

(1) 3^x X 5^y = 75
(2) 3^(x-1)(y-2) = 1


Hi Bunuel!
Looks like you made a typo in your solution! There should be 3^(x-1)(y-2) = 1 not 5^(x-1)(y-2) = 1 as in your text. Yes?
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Re: What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1  [#permalink]

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New post 03 Mar 2015, 05:42
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Re: What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1  [#permalink]

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New post 28 Jun 2016, 00:32
I can't seem to get my head around this one. Please help me understand statement 1.

-> Going by the numbers manpreetsingh86 has picked up, where x=0, there will be no combination of 5^y that will yield *exactly* 75. (I spent a lot of time with the calculator trying different numbers for y and went down to y=2.682606-2.682607). I agree we will have an approximate 75 but not an integer 75.

->Neither does Bunuel 's numbers fit in.

I had picked A because both 3 and 5 are prime numbers and any non-integer exponent will yield a non-integer to us. It will only yield an integer if we pick integer exponents. What scenario am I missing from my analysis?
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Re: What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1  [#permalink]

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New post 28 Jun 2016, 05:24
AmoyV wrote:
I can't seem to get my head around this one. Please help me understand statement 1.

-> Going by the numbers manpreetsingh86 has picked up, where x=0, there will be no combination of 5^y that will yield *exactly* 75. (I spent a lot of time with the calculator trying different numbers for y and went down to y=2.682606-2.682607). I agree we will have an approximate 75 but not an integer 75.

->Neither does Bunuel 's numbers fit in.

I had picked A because both 3 and 5 are prime numbers and any non-integer exponent will yield a non-integer to us. It will only yield an integer if we pick integer exponents. What scenario am I missing from my analysis?


You are missing the scenario with irrational numbers.

\(3^x*5^y=75\). If \(y=1\) then \(3^x*5^y=3^x*5=75\) --> \(3^x=25\) --> \(x=some \ irrational \ #\approx{2.9}\)). So, there is some irrational number x, for which 3^x = 25.
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Re: What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1  [#permalink]

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New post 01 Jul 2016, 02:16
Bunuel wrote:
AmoyV wrote:
I can't seem to get my head around this one. Please help me understand statement 1.

-> Going by the numbers manpreetsingh86 has picked up, where x=0, there will be no combination of 5^y that will yield *exactly* 75. (I spent a lot of time with the calculator trying different numbers for y and went down to y=2.682606-2.682607). I agree we will have an approximate 75 but not an integer 75.

->Neither does Bunuel 's numbers fit in.

I had picked A because both 3 and 5 are prime numbers and any non-integer exponent will yield a non-integer to us. It will only yield an integer if we pick integer exponents. What scenario am I missing from my analysis?


You are missing the scenario with irrational numbers.

\(3^x*5^y=75\). If \(y=1\) then \(3^x*5^y=3^x*5=75\) --> \(3^x=25\) --> \(x=some \ irrational \ #\approx{2.9}\)). So, there is some irrational number x, for which 3^x = 25.



Bunuel - Can we safely assume that there will an irrational number which will make 3^x equal to 25. Indeed, I too could not locate any number till 5 decimal places.
Is it safe to make such an assumption?
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Re: What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1  [#permalink]

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New post 01 Jul 2016, 03:14
Keats wrote:
Bunuel wrote:
AmoyV wrote:
I can't seem to get my head around this one. Please help me understand statement 1.

-> Going by the numbers manpreetsingh86 has picked up, where x=0, there will be no combination of 5^y that will yield *exactly* 75. (I spent a lot of time with the calculator trying different numbers for y and went down to y=2.682606-2.682607). I agree we will have an approximate 75 but not an integer 75.

->Neither does Bunuel 's numbers fit in.

I had picked A because both 3 and 5 are prime numbers and any non-integer exponent will yield a non-integer to us. It will only yield an integer if we pick integer exponents. What scenario am I missing from my analysis?


You are missing the scenario with irrational numbers.

\(3^x*5^y=75\). If \(y=1\) then \(3^x*5^y=3^x*5=75\) --> \(3^x=25\) --> \(x=some \ irrational \ #\approx{2.9}\)). So, there is some irrational number x, for which 3^x = 25.



Bunuel - Can we safely assume that there will an irrational number which will make 3^x equal to 25. Indeed, I too could not locate any number till 5 decimal places.
Is it safe to make such an assumption?


This is not an assumption: 3^x = 25 has a solution for x: \(x=\frac{2log(5)}{log(3)} \approx 2.929947...\)
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Re: What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1  [#permalink]

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Re: What is the value of x ? (1) 3^x X 5^y = 75 (2) 3^(x-1)(y-2) = 1   [#permalink] 16 Jul 2017, 00:27
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