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NEW!!! Tough and tricky exponents and roots questions
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Bunuel
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NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  Updated on: 15 Mar 2019, 03:10
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Exponents and roots problems are very common on the GMAT. So, it's extremely important to know how to manipulate them, how to factor out, take roots, multiply, divide, etc. Below are 11 problems to test your skills. Please post your thought process/solutions along with the answers.

I'll post OA's with detailed solutions tomorrow. Good luck.

1. What is the value of \(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}}\)?
A. \(2\sqrt{5}\)
B. \(\sqrt{55}\)
C. \(2\sqrt{15}\)
D. 50
E. 60

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029216

2. What is the units digit of \((17^3)^4-1973^{3^2}\)?
A. 0
B. 2
C. 4
D. 6
E. 8

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029219

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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029221

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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029222

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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029223

6. If \(x=\sqrt[5]{-37}\) then which of the following must be true?
A. \(\sqrt{-x}>2\)
B. x>-2
C. x^2<4
D. x^3<-8
E. x^4>32

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029224

7. If \(x=\sqrt{10}+\sqrt[3]{9}+\sqrt[4]{8}+\sqrt[5]{7}+\sqrt[6]{6}+\sqrt[7]{5}+\sqrt[8]{4}+\sqrt[9]{3}+\sqrt[10]{2}\), then which of the following must be true:
A. x<6
B. 6<x<8
C. 8<x<10
D. 10<x<12
E. x>12

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029227

8. If \(x\) is a positive number and equals to \(\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}\), where the given expression extends to an infinite number of roots, then what is the value of x?
A. \(\sqrt{6}\)
B. 3
C. \(1+\sqrt{6}\)
D. \(2\sqrt{3}\)
E. 6

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029228

9. If \(x\) is a positive integer then the value of \(\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}\) is closest to which of the following?
A. \(2^{11x}\)
B. \(11^{11x}\)
C. \(22^{11x}\)
D. \(2^{22x}*11^{11x}\)
E. \(2^{22x}*11^{22x}\)

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029229

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

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029231

11. If \(x>0\), \(x^2=2^{64}\) and \(x^x=2^y\) then what is the value of \(y\)?
A. 2
B. 2^(11)
C. 2^(32)
D. 2^(37)
E. 2^(64)

Solution: tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029232
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Originally posted by Bunuel on 12 Jan 2012, 02:03.
Last edited by Bunuel on 15 Mar 2019, 03:10, edited 1 time in total.
Updated.
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gmatrant
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  13 Jan 2013, 23:38
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
gmatrant
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  14 Jan 2013, 00:20
Expert Reply
gmatrant wrote:
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
gmatrant


First of all 757,571 - 58,299,374,483 = -58,298,616,912

You could consider easier cases: 11-13=-2, 11-23=-12, 21-83=-62, ...
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  30 Mar 2013, 13:08
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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] New post  31 Mar 2013, 07:33
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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] New post  16 Apr 2013, 21:53
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
H
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  16 Apr 2013, 23:09
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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] New post  17 Apr 2013, 15:17
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...

Archit
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  17 Apr 2013, 15:20
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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] New post  20 Jul 2013, 06:44
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] New post  20 Jul 2013, 06:50
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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] New post  20 Jul 2013, 07:15
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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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  20 Jul 2013, 08:14
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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] New post  22 Jul 2013, 08:28
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] New post  22 Jul 2013, 08:39
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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] New post  17 Dec 2013, 18:51
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] New post  18 Dec 2013, 01:04
1
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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] New post  11 Jun 2014, 00:46
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] New post  11 Jun 2014, 00:59
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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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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  11 Jun 2014, 01:09
\((2^{\sqrt{y}})^2=25^2\)

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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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink] New post  11 Jun 2014, 01:11
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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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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]
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