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# NEW!!! Tough and tricky exponents and roots questions

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

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12 Jan 2012, 03:03
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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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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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30 Mar 2013, 14:08
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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31 Mar 2013, 08:33
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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]

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16 Apr 2013, 22:34
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
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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

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17 Apr 2013, 00:09
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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]

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17 Apr 2013, 00:10
imhimanshu wrote:
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

|5|=5, not 5 or -5, absolute value cannot be negative.

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

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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...

Archit

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

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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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30 Apr 2013, 07:28
Bunuel wrote:
So, we have that the units digit of $$(17^3)^4=17^{12}$$ is 1 and the units digit of $$1973^3^2=1973^9$$ 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 $$(17^3)^4-1973^{3^2}$$ is 2.

Answer B.

Hi Experts,

Can you please clarify the above Red part. how can we make sure, without finding the last two digits, that the value will be 11-13. It could have been possible that the value may be something like- 21-13(i.e it is not necessary that the tens digit of smaller number is smaller or equal to the tens digit of bigger number). though I understand that the result must be negative.
Please explain.

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

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30 Apr 2013, 07:41
imhimanshu wrote:
Bunuel wrote:
So, we have that the units digit of $$(17^3)^4=17^{12}$$ is 1 and the units digit of $$1973^3^2=1973^9$$ 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 $$(17^3)^4-1973^{3^2}$$ is 2.

Answer B.

Hi Experts,

Can you please clarify the above Red part. how can we make sure, without finding the last two digits, that the value will be 11-13. It could have been possible that the value may be something like- 21-13(i.e it is not necessary that the tens digit of smaller number is smaller or equal to the tens digit of bigger number). though I understand that the result must be negative.
Please explain.

Regards,
H

(positive number ending with 1) - (greater number ending with 3) = (negative number ending with 2)

11-23=-12
31-133=-102
41-123=-82
....

Hope it's clear.

Discussed here:
new-tough-and-tricky-exponents-and-roots-questions-125956-40.html#p1029219
new-tough-and-tricky-exponents-and-roots-questions-125956-60.html#p1054715
new-tough-and-tricky-exponents-and-roots-questions-125956-60.html#p1054715
new-tough-and-tricky-exponents-and-roots-questions-125956-80.html#p1168559
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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01 May 2013, 04:59
Bunuel wrote:

(positive number ending with 1) - (greater number ending with 3) = (negative number ending with 2)

11-23=-12
31-133=-102
41-123=-82
....

Thanks Bunuel, I fell into the Trap. Your explanations are really useful.

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

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27 Jun 2013, 10:41
Bunuel wrote:
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)

$$x^2=2^{64}$$ --> $$x=\sqrt{2^{64}}=2^{\frac{64}{2}}=2^{32}$$ (note that $$x=-\sqrt{2^{64}}$$ is not a valid solution as given that $$x>0$$).

Second step: $$x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}=2^y$$ --> $$y=2^{37}$$.

OR second step: $$x^x=(2^{32})^x=2^{32x}=2^y$$ --> $$y=32x$$ --> since $$x=2^{32}$$ then $$y=32x=32*2^{32}=2^5*2^{32}=2^{37}$$.

Answer: D.

Had a wonderful time solving and learning from these challenges, I really admire the way you break every equation using algebra.

BDW are these sets from GC club challenges??
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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27 Jun 2013, 10:43
prateekbhatt wrote:
Bunuel wrote:
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)

$$x^2=2^{64}$$ --> $$x=\sqrt{2^{64}}=2^{\frac{64}{2}}=2^{32}$$ (note that $$x=-\sqrt{2^{64}}$$ is not a valid solution as given that $$x>0$$).

Second step: $$x^x=(2^{32})^{(2^{32})}=2^{32*2^{32}}=2^{2^{5}*2^{32}}=2^{2^{37}}=2^y$$ --> $$y=2^{37}$$.

OR second step: $$x^x=(2^{32})^x=2^{32x}=2^y$$ --> $$y=32x$$ --> since $$x=2^{32}$$ then $$y=32x=32*2^{32}=2^5*2^{32}=2^{37}$$.

Answer: D.

Had a wonderful time solving and learning from these challenges, I really admire the way you break every equation using algebra.

BDW are these sets from GC club challenges??

Yes, almost all my questions are included in the GMAT Club Tests.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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09 Jul 2013, 00:39
Bumping for review and further discussion*. Get a kudos point for an alternative solution!

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

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09 Jul 2013, 11:31
BDSunDevil wrote:
Edited
What is the value of 5+4*5+4*5^2+4*5^3+4*5^4+4*5^5?
= 5+4*5[1+5+5^2+5^3+5^4]
=5+20*781
=15625
=625*25=5^6

this can also be solved as below:

$$5 + 4 * (5 + 5^2 + 5^3 + 5^4 + 5^5)$$

the highlighted series is a GP series....

==> sum of the items in GP is = $$a ( r^n -1)/(r-1)$$

total = $$5 + 4 * 5(5^5-1)/(5-1)$$

$$= 5 + 5^6 - 5 = 5^6$$
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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09 Jul 2013, 11:57
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

let y = $$\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}$$ ==> $$y=\sqrt{6+y}$$
==> y^2 = 6+y
resolve this equation to get roots of +3 and -2. Since x is a positive number, x = 3
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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

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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] 20 Jul 2013, 09:14

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