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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, 02:03
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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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13 Jan 2012, 20:46
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Q4. What is the value of $$5+4*5+4*5^2+4*5^3+4*5^4+4*5^5$$

This one was actually the simplest I thought. Here is how:

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

So $$5*(1+4)+4*5^2+4*5^3+4*5^4+4*5^5$$
So $$5^2+4*5^2+4*5^3+4*5^4+4*5^5$$
So $$5^2*(1+4)+4*5^3+4*5^4+4*5^5$$
So $$5^3+4*5^3+4*5^4+4*5^5$$
So $$5^3*(1+4)+4*5^4+4*5^5$$

So every expression behind contributes a power of 1 to the one in front of it. We just need to see the last which is $$5^5$$
Keep solving and you come to a total of $$5^6$$

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

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14 Jan 2012, 10:30
IanStewart wrote:
Your substitution might be making things more difficult. If we want to estimate the value of

$$\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}$$

then we can ignore the comparatively small values in the numerator and denominator; this is roughly equal to

$$\frac{22^{22x}}{11^{11x}}$$

which is equal to

$$\frac{22^{22x}}{11^{11x}} = \frac{2^{22x} 11^{22x}}{11^{11x}} = 2^{22x} 11^{11x}$$

ah, I got what u meant (at last! heh)
so, this q. is similar to the q "2001/6001 is close to which of the following?" the answ- 2/6=1/3
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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14 Jan 2012, 10:44
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SonyGmat wrote:
LalaB wrote:
10. Given that 5x=125-3y+z and \sqrt{5x}-25-\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

5x-125=z-3y

sqrt{5x}-25=\sqrt{z-3y}
(sqrt{5x}-25)^2=z-3y

(sqrt{5x}-25)^2=5x-125
\sqrt{5x}=15
x=5*9=45

so the expression \sqrt{\frac{45(z-3y)}{x} = \sqrt{45(15-25)/45}=\sqrt{-10}

E is the answ

bunuel wrote:
10. Given that $$5x=125-3y+z$$ and $$\sqrt{5x}-25-\sqrt{z-3y}=0$$, 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

We know that:

$$5x=125-3y+z$$ ---> [highlight]$$z-3y=5x-125$$ (1)[/highlight]

and that:

$$\sqrt{5x}-25-\sqrt{z-3y}=0$$ ---> $$\sqrt{5x}-25=\sqrt{z-3y}$$ ----> $$[\sqrt{5x}-25]^2=[\sqrt{z-3y}]^2$$ ----> using (1): $$[\sqrt{5x}-25]^2=[\sqrt{5x-125}]^2$$ --->

$$5x-50\sqrt{5x}+625=5x-125$$ ---> $$50\sqrt{5x}=750$$ ---> $$\sqrt{5x}=15$$ ---> $$5x=15*15$$--->[[highlight]$$x=45$$ (2)[/highlight]

(1), (2) ---> $$z-3y=5*45-125$$--->[highlight]$$z-3y=100$$ (3)[/highlight]

therefore: $$\sqrt{\frac{45(z-3y)}{x}}$$ ---->using (2),(3) :$$\sqrt{\frac{45*100}{45}}$$ ---> 10 ---->B

hm, strange enough. I seem to overtaxed myself . i referred to the 2nd equation, since we need to find sqroot(z-3y)
\sqrt{5x}-25=\sqrt{z-3y}
\sqrt{5*45}-25=\sqrt{z-3y}
-10=\sqrt{z-3y}

but at the same time 5x=125-3y+z 5*45-125=z-3y 10=z-3y
it seems there is a typo in a q stem -\sqrt{5x}-25-\sqrt{z-3y}=0

it must be \sqrt{5x}-5-\sqrt{z-3y}=0
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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14 Jan 2012, 11:21
LalaB wrote:
SonyGmat wrote:
LalaB wrote:
10. Given that 5x=125-3y+z and \sqrt{5x}-25-\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

5x-125=z-3y

sqrt{5x}-25=\sqrt{z-3y}
(sqrt{5x}-25)^2=z-3y

(sqrt{5x}-25)^2=5x-125
\sqrt{5x}=15
x=5*9=45

so the expression \sqrt{\frac{45(z-3y)}{x} = \sqrt{45(15-25)/45}=\sqrt{-10}

E is the answ

bunuel wrote:
10. Given that $$5x=125-3y+z$$ and $$\sqrt{5x}-25-\sqrt{z-3y}=0$$, 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

We know that:

$$5x=125-3y+z$$ ---> [highlight]$$z-3y=5x-125$$ (1)[/highlight]

and that:

$$\sqrt{5x}-25-\sqrt{z-3y}=0$$ ---> $$\sqrt{5x}-25=\sqrt{z-3y}$$ ----> $$[\sqrt{5x}-25]^2=[\sqrt{z-3y}]^2$$ ----> using (1): $$[\sqrt{5x}-25]^2=[\sqrt{5x-125}]^2$$ --->

$$5x-50\sqrt{5x}+625=5x-125$$ ---> $$50\sqrt{5x}=750$$ ---> $$\sqrt{5x}=15$$ ---> $$5x=15*15$$--->[[highlight]$$x=45$$ (2)[/highlight]

(1), (2) ---> $$z-3y=5*45-125$$--->[highlight]$$z-3y=100$$ (3)[/highlight]

therefore: $$\sqrt{\frac{45(z-3y)}{x}}$$ ---->using (2),(3) :$$\sqrt{\frac{45*100}{45}}$$ ---> 10 ---->B

hm, strange enough. I seem to overtaxed myself . i referred to the 2nd equation, since we need to find sqroot(z-3y)
\sqrt{5x}-25=\sqrt{z-3y}
\sqrt{5*45}-25=\sqrt{z-3y}
-10=\sqrt{z-3y}

but at the same time 5x=125-3y+z 5*45-125=z-3y 10=z-3y
it seems there is a typo in a q stem -\sqrt{5x}-25-\sqrt{z-3y}=0

it must be \sqrt{5x}-5-\sqrt{z-3y}=0

Yes, there is a typo: must be 5 instead of 25. Thanks. Edited.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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14 Jan 2012, 13:53
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SOLUTIONS:

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

Square the given expression to get rid of the roots, though don't forget to un-square the value you get at the end to balance this operation and obtain the right answer:

Must know fro the GMAT: $$(x+y)^2=x^2+2xy+y^2$$ (while $$(x-y)^2=x^2-2xy+y^2$$).

So we get: $$(\sqrt{25+10\sqrt{6}}+\sqrt{25-10\sqrt{6}})^2=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(\sqrt{25-10\sqrt{6}})^2=$$
$$=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})+(25-10\sqrt{6})$$.

Note that sum of the first and the third terms simplifies to $$(25+10\sqrt{6})+(25-10\sqrt{6})=50$$, so we have $$50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})$$ --> $$50+2(\sqrt{25+10\sqrt{6}})(\sqrt{25-10\sqrt{6}})=50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}$$.

Also must know for the GMAT: $$(x+y)(x-y)=x^2-y^2$$, thus $$50+2\sqrt{(25+10\sqrt{6})(25-10\sqrt{6})}=50+2\sqrt{25^2-(10\sqrt{6})^2)}=50+2\sqrt{625-600}=50+2\sqrt{25}=60$$.

Recall that we should un-square this value to get the right the answer: $$\sqrt{60}=2\sqrt{15}$$.

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

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14 Jan 2012, 13:58
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2. What is the units digit of $$(17^3)^4-1973^{3^2}$$?
A. 0
B. 2
C. 4
D. 6
E. 8

Must know for the GMAT:
I. The units digit of $$(abc)^n$$ is the same as that of $$c^n$$, which means that the units digit of $$(17^3)^4$$ is that same as that of $$(7^3)^4$$ and the units digit of $$1973^{3^2}$$ is that same as that of $$3^{3^2}$$.

II. If exponentiation is indicated by stacked symbols, the rule is to work from the top down, thus:
$$a^m^n=a^{(m^n)}$$ and not $$(a^m)^n$$, which on the other hand equals to $$a^{mn}$$.

So:
$$(a^m)^n=a^{mn}$$;

$$a^m^n=a^{(m^n)}$$.

Thus, $$(7^3)^4=7^{(3*4)}=7^{12}$$ and $$3^{3^2}=3^{(3^2)}=3^9$$.

III. The units digit of integers in positive integer power repeats in specific pattern (cyclicity): The units digit of 7 and 3 in positive integer power repeats in patterns of 4:

1. 7^1=7 (last digit is 7)
2. 7^2=9 (last digit is 9)
3. 7^3=3 (last digit is 3)
4. 7^4=1 (last digit is 1)

5. 7^5=7 (last digit is 7 again!)
...

1. 3^1=3 (last digit is 3)
2. 3^2=9 (last digit is 9)
3. 3^3=27 (last digit is 7)
4. 3^4=81 (last digit is 1)

5. 3^5=243 (last digit is 3 again!)
...

Thus th units digit of $$7^{12}$$ will be 1 (4th in pattern, as 12 is a multiple of cyclicty number 4) and the units digit of $$3^9$$ will be 3 (first in pattern, as 9=4*2+1).

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.

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14 Jan 2012, 14:00
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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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14 Jan 2012, 14:03
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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.

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14 Jan 2012, 14:05
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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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14 Jan 2012, 14:06
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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

Must know for the GMAT: Even roots from negative number is undefined on the GMAT (as GMAT is dealing only with Real Numbers): $$\sqrt[{even}]{negative}=undefined$$, for example $$\sqrt{-25}=undefined$$.

Odd roots have the same sign as the base of the root. For example, $$\sqrt[3]{125} =5$$ and $$\sqrt[3]{-64} =-4$$.

Back to the original question:

As $$-2^5=-32$$ then $$x$$ must be a little bit less than -2 --> $$x=\sqrt[5]{-37}\approx{-2.1}<-2$$. Thus $$x^3\approx{(-2.1)^3}\approx{-8.something}<-8$$, so option D must be true.

As for the other options:
A. $$\sqrt{-x}=\sqrt{-(-2.1)}=\sqrt{2.1}<2$$, $$\sqrt{-x}>2$$ is not true.
B. $$x\approx{-2.1}<-2$$, thus x>-2 is also not true.
C. $$x^2\approx{(-2.1)}^2=4.something>4$$, thus x^2<4 is also not true.
E. $$x^4\approx{(-2.1)}^4\approx17$$, (2^4=16, so anyway -2.1^4 can not be more than 32) thus x^4>32 is also not true.

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

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14 Jan 2012, 14:08
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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

Here is a little trick: any positive integer root from a number more than 1 will be more than 1. For example: $$\sqrt[1000]{2}>1$$.

Now, $$\sqrt{10}>3$$ (as 3^2=9) and $$\sqrt[3]{9}>2$$ (2^3=8). Thus $$x=(# \ more \ then \ 3)+(# \ more \ then \ 2)+(7 \ numbers \ more \ then \ 1)=$$
$$=(# \ more \ then \ 5)+(# \ more \ then \ 7)=$$
$$=(# \ more \ then \ 12)$$

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14 Jan 2012, 14:09
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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

Given: $$x>0$$ and $$x=\sqrt{6+{\sqrt{6+\sqrt{6+\sqrt{6+...}}}}}$$ --> $$x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}$$, as the expression under the square root extends infinitely then expression in brackets would equal to $$x$$ itself and we can safely replace it with $$x$$ and rewrite the given expression as $$x=\sqrt{6+x}$$. Square both sides: $$x^2=6+x$$ --> $$(x+2)(x-3)=0$$ --> $$x=-2$$ or $$x=3$$, but since $$x>0$$ then: $$x=3$$.

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14 Jan 2012, 14:11
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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. $$2^{22x}*11^{22x}$$

Note that we need approximate value of the given expression. Now, $$22^{22x}$$ is much larger number than $$22^{2x}$$. Hence $$22^{22x}-22^{2x}$$ will be very close to $$22^{22x}$$ itself, basically $$22^{2x}$$ is negligible in this case. The same way $$11^{11x}-11^x$$ will be very close to $$11^{11x}$$ itself.

Thus $$\frac{22^{22x}-22^x}{11^{11x}-11^x}\approx{\frac{22^{22x}}{11^{11x}}}=\frac{2^{22x}*11^{22x}}{11^{11x}}=2^{22x}*11^{11x}$$.

You can check this algebraically as well: $$\frac{22^{22x}-22^{2x}}{11^{11x}-11^x}=\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}$$. Again, -1, both in denominator and nominator is negligible value and we'll get the same expression as above: $$\frac{22^{2x}(22^{20x}-1)}{11^x(11^{10x}-1)}\approx{\frac{22^{2x}*22^{20x}}{11^x*11^{10x}}}=\frac{22^{22x}}{11^{11x}}=2^{22x}*11^{11x}$$

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

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14 Jan 2012, 14:13
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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, $$\sqrt{\frac{45(z-3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10$$.

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14 Jan 2012, 14:15
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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 $$y=32x=32*2^{32}=2^5*2^{32}=2^{37}$$.

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

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14 Jan 2012, 14:28
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Just posted solutions. Kudos points given to everyone with correct solutions. Let me know if I missed someone.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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15 Jan 2012, 09:23
So the king is back with a bang
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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15 Jan 2012, 19:59
Bunuel, thank you for these tough questions. It was interesting to solve them.
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Re: NEW!!! Tough and tricky exponents and roots questions [#permalink]

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15 Jan 2012, 20:04
“The only true wisdom is in knowing you know nothing.”
― Socrates

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

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22 Jan 2012, 03:19
Wow. Brunei, what did you get on your GMAT?
Re: NEW!!! Tough and tricky exponents and roots questions   [#permalink] 22 Jan 2012, 03:19

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

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