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NEW!!! Tough and tricky exponents and roots questions

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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 14 Jan 2013, 00:38
Thanks Bunuel for these interesting problems.

With respect to problem number 2

"So, we have that the units digit of is 1 and the units digit of is 3. Also notice that the second number is much larger then the first one, thus their difference will be negative, something like 11-13=-2, which gives the final answer that the units digit of is 2."

I don't understand the above statement. If the last digit is 1 for the smaller number and the last digit is 3 for the larger number.
And we are trying to find different between smaller number - larger number

eg 757571 - 58299374483. We borrow one from the previous digit and make 1 as 11 and then subtract 3 from 11. So shouldn't the last digit be 8.

I am unable to think any other way. Please let me know where I am going wrong.

Thanks
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 30 Mar 2013, 14:08
1
You talk about even and odd in this problem. Wouldn't this not be enough? Since 26 has factors of 2 and 13, even if you had added some even numbers in the product it would still not have sufficed? You would need a 13, 39 etc ? Is my reasoning correct?

Bunuel wrote:
5. If \(x=23^2*25^4*27^6*29^8\) and is a multiple of \(26^n\), where \(n\) is a non-negative integer, then what is the value of \(n^{26}-26^n\)?
A. -26
B. -25
C. -1
D. 0
E. 1

\(23^2*25^4*27^6*29^8=odd*odd*odd*odd=odd\) so \(x\) is an odd number. The only way it to be a multiple of \(26^n\) (even number in integer power) is when \(n=0\), in this case \(26^n=26^0=1\) and 1 is a factor of every integer. Thus \(n=0\) --> \(n^{26}-26^n=0^{26}-26^0=0-1=-1\). Must know for the GMAT: \(a^0=1\), for \(a\neq{0}\) - any nonzero number to the power of 0 is 1. Important note: the case of 0^0 is not tested on the GMAT.

Answer: C.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 31 Mar 2013, 08:33
jeremystaub28 wrote:
You talk about even and odd in this problem. Wouldn't this not be enough? Since 26 has factors of 2 and 13, even if you had added some even numbers in the product it would still not have sufficed? You would need a 13, 39 etc ? Is my reasoning correct?

Bunuel wrote:
5. If \(x=23^2*25^4*27^6*29^8\) and is a multiple of \(26^n\), where \(n\) is a non-negative integer, then what is the value of \(n^{26}-26^n\)?
A. -26
B. -25
C. -1
D. 0
E. 1

\(23^2*25^4*27^6*29^8=odd*odd*odd*odd=odd\) so \(x\) is an odd number. The only way it to be a multiple of \(26^n\) (even number in integer power) is when \(n=0\), in this case \(26^n=26^0=1\) and 1 is a factor of every integer. Thus \(n=0\) --> \(n^{26}-26^n=0^{26}-26^0=0-1=-1\). Must know for the GMAT: \(a^0=1\), for \(a\neq{0}\) - any nonzero number to the power of 0 is 1. Important note: the case of 0^0 is not tested on the GMAT.

Answer: C.


The fact that the product is odd, is already enough to say that it cannot be a multiple of 26^n unless n=0.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 16 Apr 2013, 22:53
Bunuel wrote:

• \(\sqrt{x^2}=|x|\), when \(x\leq{0}\), then \(\sqrt{x^2}=-x\) and when \(x\geq{0}\), then \(\sqrt{x^2}=x\)

• When the GMAT provides the square root sign for an even root, such as \(\sqrt{x}\) or \(\sqrt[4]{x}\), then the only accepted answer is the positive root.

That is, \(\sqrt{25}=5\), NOT +5 or -5. In contrast, the equation \(x^2=25\) has TWO solutions, +5 and -5. Even roots have only a positive value on the GMAT.



Hi Bunuel,

I am well versed with these exponent rules, however I am becoming confused with the above two rules stated-

If \(\sqrt{x^2}\) =\(|x|\) then -> \(\sqrt{5^2}\) = \(|5|\), which means that value can be either 5 or -5. However, as per your example, value should be 5 only. How come this is true? Please clarify.

Thanks
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New post 17 Apr 2013, 00:09
1
imhimanshu wrote:
Bunuel wrote:
10. Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined



Hi Bunuel,

Request you to please look at my below solution and kindly guide where I am going wrong.-

Given:
5x = 125-3y +z ----(1)

\(\sqrt{5x} = 5+\sqrt{z-3y}\)
Squaring both sides

\(5x= (5+\sqrt{z-3y})^2\)

\(5x= 125+(z-3y) +10\sqrt{(z-3y)}\) ---------------(2)

If you compare equation 1 and 2; then the term \(\sqrt{(z-3y)}\) has to be 0

If I put this value in the \(\sqrt{\frac{45(z-3y)}{x}}\)
Then, the value comes out to be \(\sqrt{0}.\). This is Absurd.

I know this is incorrect, but would like to know where I am going wrong.

Your help will be appreciated.

Thanks
H


When you square \(5x= (5+\sqrt{z-3y})^2\) you should get 5x=25+10(z-3y)^1/2+(z-3y) (5^2=25 not 125).
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 17 Apr 2013, 16:17
Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14

First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\)

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.


Hi Bunuel
Can u pls explain the step.....(5^(5x))^2 = 70^2 implies 5^5x = 70
does that mean if a^n = b^n, a=b...

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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 17 Apr 2013, 16:20
Archit143 wrote:
Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14

First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\)

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.


Hi Bunuel
Can u pls explain the step.....(5^(5x))^2 = 70^2 implies 5^5x = 70
does that mean if a^n = b^n, a=b...

Archit


It's very simple, take the square root from \((5^{5x})^2=70^2\) to get \(5^{5x}=70\).
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 20 Jul 2013, 07:44
Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14

First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\)

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.


Bunuel, could you please kindly explain how you come up with the 70 here? Many thanks
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 20 Jul 2013, 07:50
BankerRUS wrote:
Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14

First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\)

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.


Bunuel, could you please kindly explain how you come up with the 70 here? Many thanks


Can you please tell me which step is unclear:

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\).
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 20 Jul 2013, 08:15
Bunuel wrote:
BankerRUS wrote:
Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14

First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\)

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.


Bunuel, could you please kindly explain how you come up with the 70 here? Many thanks


Can you please tell me which step is unclear:

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\).


The first step with the 4,900 is clear. I do not understand the following: 5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14[/m]

and how you get to the 70 there.
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New post 20 Jul 2013, 09:14
BankerRUS wrote:
Bunuel wrote:
BankerRUS wrote:

Bunuel, could you please kindly explain how you come up with the 70 here? Many thanks


Can you please tell me which step is unclear:

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\).


The first step with the 4,900 is clear. I do not understand the following: 5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14[/m]

and how you get to the 70 there.


\(5^{5x}=70\). So, this step is clear.

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2\). Is this step clear?

Next, replace \(5^{5x}\) with 70 and \(2^{\sqrt{y}}\) with 25: \(70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\).

Does this make sense?
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 22 Jul 2013, 09:28
Edit: [edit]
[/edit]
Bunuel wrote:
4. What is the value of \(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5\)?
A. 5^6
B. 5^7
C. 5^8
D. 5^9
E. 5^10

This question can be solved in several ways:

Traditional approach: \(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5=5+4(5+5^2+5^3+5^4+5^5)\) Note that we have the sum of geometric progression in brackets with first term equal to 5 and common ratio also equal to 5. The sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}-1)}{r-1}\), where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\).

So in our case: \(5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^5-1)}{5-1})=5^6\).

30 sec approach based on answer choices:
We have the sum of 6 terms. Now, if all terms were equal to the largest term 4*5^5 we would have: \(sum=6*(4*5^5)=24*5^5\approx{5^2*5^5}\approx{5^7}\), so the actual sum must be less than 5^7, thus the answer must be A: 5^6.

Answer: A.



Thanks for the explanation. The part where you take 5+4 out of the full equation is still unclear to me - 5+4(5+5^2+5^3+5^4+5^5). Could you please break it down into multiple steps?
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 22 Jul 2013, 09:39
vaishnogmat wrote:
Edit: [edit]
[/edit]
Bunuel wrote:
4. What is the value of \(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5\)?
A. 5^6
B. 5^7
C. 5^8
D. 5^9
E. 5^10

This question can be solved in several ways:

Traditional approach: \(5+4*5+4*5^2+4*5^3+4*5^4+4*5^5=5+4(5+5^2+5^3+5^4+5^5)\) Note that we have the sum of geometric progression in brackets with first term equal to 5 and common ratio also equal to 5. The sum of the first \(n\) terms of geometric progression is given by: \(sum=\frac{b*(r^{n}-1)}{r-1}\), where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\).

So in our case: \(5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^5-1)}{5-1})=5^6\).

30 sec approach based on answer choices:
We have the sum of 6 terms. Now, if all terms were equal to the largest term 4*5^5 we would have: \(sum=6*(4*5^5)=24*5^5\approx{5^2*5^5}\approx{5^7}\), so the actual sum must be less than 5^7, thus the answer must be A: 5^6.

Answer: A.



Thanks for the explanation. The part where you take 5+4 out of the full equation is still unclear to me - 5+4(5+5^2+5^3+5^4+5^5). Could you please break it down into multiple steps?


4 is factored out from 4*5+4*5^2+4*5^3+4*5^4+4*5^5, not 5+4 from the whole expression: \(4*5+4*5^2+4*5^3+4*5^4+4*5^5=4(5+5^2+5^3+5^4+5^5)\).

So, \(5+(4*5+4*5^2+4*5^3+4*5^4+4*5^5)=5+4(5+5^2+5^3+5^4+5^5)\).

Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 17 Dec 2013, 19:51
Bunuel wrote:
10. Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined

Rearranging both expressions we'll get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have that \(a^2-b^2=125\) and \(a-b=5\). Now, \(a^2-b^2=(a-b)(a+b)=125\) and as \(a-b=5\) then \((a-b)(a+b)=5*(a+b)=125\) --> \(a+b=25\). Thus we get two equations with two unknowns: \(a+b=25\) and \(a-b=5\) --> solving for \(a\) --> \(a=15=\sqrt{5x}\) --> \(x=45\). Solving for \(b\) -->\(b=10=\sqrt{z-3y}\)

Finally, \(\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10\).

Answer: B.


This is how I went about solving this problem:

Given \(\sqrt{5x} - 5 - \sqrt{z-3y}= 0\):

Rearrange and square
\((\sqrt{5x})^2 = (\sqrt{z-3y}+5)^2\)
\(|5x| = z-3y + 25 + 10\sqrt{z-3y}\)

Assuming that x is positive and given \(5x=125-3y+z\), set the two equations equal to each other
\(125 - 3y + z = z-3y + 25 + 10\sqrt{z-3y}\)
\(100 = 10\sqrt{z-3y}\)
\(100 = z-3y\)

\(5x = 125-3y+z\)
\(5x = 125 + 100\)
\(x = 45\)

Thus by substitution:
\(= \sqrt{\frac{45(z-3y)}{x}}\)

\(= \sqrt{\frac{45(100)}{45}}\)

\(= \sqrt{100}\)

\(= 10\)

However, notice that I had to assume that |5x| = 5x
Why is this permissible in this problem?
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 18 Dec 2013, 02:04
1
NateTheGreat11 wrote:
Bunuel wrote:
10. Given that \(5x=125-3y+z\) and \(\sqrt{5x}-5-\sqrt{z-3y}=0\), then what is the value of \(\sqrt{\frac{45(z-3y)}{x}}\)?
A. 5
B. 10
C. 15
D. 20
E. Can not be determined

Rearranging both expressions we'll get: \(5x-(z-3y)=125\) and \(\sqrt{5x}-\sqrt{z-3y}=5\). Denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z-3y}\) as \(b\).

So we have that \(a^2-b^2=125\) and \(a-b=5\). Now, \(a^2-b^2=(a-b)(a+b)=125\) and as \(a-b=5\) then \((a-b)(a+b)=5*(a+b)=125\) --> \(a+b=25\). Thus we get two equations with two unknowns: \(a+b=25\) and \(a-b=5\) --> solving for \(a\) --> \(a=15=\sqrt{5x}\) --> \(x=45\). Solving for \(b\) -->\(b=10=\sqrt{z-3y}\)

Finally, \(\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10\).

Answer: B.


This is how I went about solving this problem:

Given \(\sqrt{5x} - 5 - \sqrt{z-3y}= 0\):

Rearrange and square
\((\sqrt{5x})^2 = (\sqrt{z-3y}+5)^2\)
\(|5x| = z-3y + 25 + 10\sqrt{z-3y}\)

Assuming that x is positive and given \(5x=125-3y+z\), set the two equations equal to each other
\(125 - 3y + z = z-3y + 25 + 10\sqrt{z-3y}\)
\(100 = 10\sqrt{z-3y}\)
\(100 = z-3y\)

\(5x = 125-3y+z\)
\(5x = 125 + 100\)
\(x = 45\)

Thus by substitution:
\(= \sqrt{\frac{45(z-3y)}{x}}\)

\(= \sqrt{\frac{45(100)}{45}}\)

\(= \sqrt{100}\)

\(= 10\)

However, notice that I had to assume that |5x| = 5x
Why is this permissible in this problem?


Even roots from negative numbers are undefined for the GMAT. We have \(\sqrt{5x}\) in one of the expressions, which means that x is non-negative, thus |5x|=5x.

Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 11 Jun 2014, 01:46
5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2

How come you arrive to 25 from y?

Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14

First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\)

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.

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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 11 Jun 2014, 01:59
arindamsur wrote:
5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2

How come you arrive to 25 from y?

Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14

First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\)

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.


\(4^{\sqrt{y}}=(2^2)^{\sqrt{y}}=2^{2*\sqrt{y}}=(2^{\sqrt{y}})^2=25^2\) --> .

P.S. Please go this post: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum). Thank you.
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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.


What are GMAT Club Tests?
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 11 Jun 2014, 02:09
\((2^{\sqrt{y}})^2=25^2\)

Bunuel explain this part please

Bunuel wrote:
arindamsur wrote:
5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2

How come you arrive to 25 from y?

Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14

First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\)

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.


\(4^{\sqrt{y}}=(2^2)^{\sqrt{y}}=2^{2*\sqrt{y}}=(2^{\sqrt{y}})^2=25^2\) --> .

P.S. Please go this post: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum). Thank you.

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Arindam Sur
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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New post 11 Jun 2014, 02:11
arindamsur wrote:
\((2^{\sqrt{y}})^2=25^2\)

Bunuel explain this part please

Bunuel wrote:

Bunuel wrote:
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}\)?
A. 14/5
B. 5
C. 28/5
D. 13
E. 14

First thing one should notice here is that \(x\) and \(y\) must be some irrational numbers (4,900 has other primes then 5 in its prime factorization and 25 doesn't have 2 as a prime at all), so we should manipulate with given expressions rather than to solve for x and y.

\(5^{10x}=4,900\) --> \((5^{5x})^2=70^2\) --> \(5^{5x}=70\)

\(\frac{(5^{(x-1)})^5}{4^{-\sqrt{y}}}=5^{(5x-5)}*4^{\sqrt{y}}=5^{5x}*5^{-5}*(2^{\sqrt{y}})^2=70*5^{-5}*25^2=70*5^{-5}*5^4=70*5^{-1}=\frac{70}{5}=14\)

Answer: E.


\(4^{\sqrt{y}}=(2^2)^{\sqrt{y}}=2^{2*\sqrt{y}}=(2^{\sqrt{y}})^2=25^2\) --> .

P.S. Please go this post: rules-for-posting-please-read-this-before-posting-133935.html#p1096628 (Writing Mathematical Formulas on the Forum). Thank you.


We are given in the stem that \(2^{\sqrt{y}}=25\), thus \((2^{\sqrt{y}})^2=25^2\).
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New to the Math Forum?
Please read this: Ultimate GMAT Quantitative Megathread | All You Need for Quant | PLEASE READ AND FOLLOW: 12 Rules for Posting!!!

Resources:
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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.


What are GMAT Club Tests?
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