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

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Joined: 13 Jun 2010
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Re: NEW!!! Tough and tricky exponents and roots questions  [#permalink]

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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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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.

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

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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.

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, 22:53
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
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.

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, 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.

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

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

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.

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

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

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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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.

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

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

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.

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

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

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.

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

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

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]

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22 Jul 2013, 09:28
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.

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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22 Jul 2013, 09:39
vaishnogmat wrote:
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.

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

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

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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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.

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

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

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

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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.

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

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

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

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

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

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.

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

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

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

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

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

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.

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

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

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

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 &nbs [#permalink] 11 Jun 2014, 02:11

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