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Re: NEW!!! Tough and tricky exponents and roots questions
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13 Jan 2012, 21:45
Q3. \(5^{10x}=4900\) so \({(5^{5x})}^2=70^2\) Take square root of both sides > so \(5^{5x}=70\) Also from question stem: \(2^{\sqrt{y}} = 5^2\) So (from the question stem) \(4^{\sqrt{y}} = \frac{1}{4^{\sqrt{y}}}\) \(4^{\sqrt{y}} = (\frac{1}{2^{\sqrt{y}}})^2 = \frac{1}{5^4}\) Now \(5^{(x1)^5}\) Should be \(\frac{5^{5x}}{5^5}\) We know \(5^{10x}=4900\) so \(5^{10x}=70^2\) so \(5^{5x}=70\) So \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}\) is basically \(\frac{5^x}{5^5}*5^4\) We already know the value of \(5^x\) which is \(70\) So now it becomes \(\frac{70}{5^5}*5^4\) Which should resolve to \(14\) Hence Answer = E = \(14\)
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14 Jan 2012, 14:58
2. What is the units digit of \((17^3)^41973^{3^2}\)?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 1113=2, which gives the final answer that the units digit of \((17^3)^41973^{3^2}\) is 2. Answer B.
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14 Jan 2012, 15:00
3. If \(5^{10x}=4,900\) and \(2^{\sqrt{y}}=25\) what is the value of \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}\)?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. \(5^{10x}=4,900\) > \((5^{5x})^2=70^2\) > \(5^{5x}=70\) \(\frac{(5^{(x1)})^5}{4^{\sqrt{y}}}=5^{(5x5)}*4^{\sqrt{y}}=5^{5x}*5^{5}*(2^{\sqrt{y}})^2=70*5^{5}*25^2=70*5^{5}*5^4=70*5^{1}=\frac{70}{5}=14\) Answer: E.
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14 Jan 2012, 15:03
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: \(sum=\frac{b*(r^{n}1)}{r1}\), where \(b\) is the first term, \(n\) # of terms and \(r\) is a common ratio \(\neq{1}\). So in our case: \(5+4(5+5^2+5^3+5^4+5^5)=5+4(\frac{5(5^51)}{51})=5^6\). 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.
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14 Jan 2012, 15:05
5. If \(x=23^2*25^4*27^6*29^8\) and is a multiple of \(26^n\), where \(n\) is a nonnegative 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 \(n=0\) > \(n^{26}26^n=0^{26}26^0=01=1\). 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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14 Jan 2012, 15:06
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. Answer: D.
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14 Jan 2012, 15:08
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 \(x=(# \ more \ then \ 3)+(# \ more \ then \ 2)+(7 \ numbers \ more \ then \ 1)=\) \(=(# \ more \ then \ 5)+(# \ more \ then \ 7)=\) \(=(# \ more \ then \ 12)\) Answer: E.
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14 Jan 2012, 15:09
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: \(x^2=6+x\) > \((x+2)(x3)=0\) > \(x=2\) or \(x=3\), but since \(x>0\) then: \(x=3\). Answer: B.
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14 Jan 2012, 15:11
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: \(\frac{22^{2x}(22^{20x}1)}{11^x(11^{10x}1)}\approx{\frac{22^{2x}*22^{20x}}{11^x*11^{10x}}}=\frac{22^{22x}}{11^{11x}}=2^{22x}*11^{11x}\) Answer: D.
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14 Jan 2012, 15:13
10. Given that \(5x=1253y+z\) and \(\sqrt{5x}5\sqrt{z3y}=0\), then what is the value of \(\sqrt{\frac{45(z3y)}{x}}\)?A. 5 B. 10 C. 15 D. 20 E. Can not be determined Rearranging both expressions we'll get: \(5x(z3y)=125\) and \(\sqrt{5x}\sqrt{z3y}=5\). Denote \(\sqrt{5x}\) as \(a\) and \(\sqrt{z3y}\) as \(b\). So we have that \(a^2b^2=125\) and \(ab=5\). Now, \(a^2b^2=(ab)(a+b)=125\) and as \(ab=5\) then \((ab)(a+b)=5*(a+b)=125\) > \(a+b=25\). Thus we get two equations with two unknowns: \(a+b=25\) and \(ab=5\) > solving for \(a\) > \(a=15=\sqrt{5x}\) > \(x=45\). Solving for \(b\) >\(b=10=\sqrt{z3y}\) Finally, \(\sqrt{\frac{45(z3y)}{x}}=\sqrt{\frac{45*10^2}{45}}=10\). Answer: B.
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14 Jan 2012, 15:15
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.
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15 Jan 2012, 21:04
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Re: NEW!!! Tough and tricky exponents and roots questions
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06 Mar 2012, 23:53
Quote: What is the units digit of (17^3)^41973^{3^2}? A. 0 B. 2 C. 4 D. 6 E. 8
Quote: 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 note that the second number is mush larger then the first one, thus their difference will be negative, something like 1113=2, which gives the final answer that the units digit of (17^3)^41973^{3^2} is 2.
Answer B. I agree with all the steps above but i have this doubt in this unit digit of 1st number is 1. Unit digit of second number is 3 13 would it not mean unit digit is 8? for eg if we have 551 853 ..Unit digit will be 2 but what if it is 551453 .unit digit will be 8? so how do we know which is the larger number of these, although it says XY? Please let me know,.



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07 Mar 2012, 01:36
shankar245 wrote: Quote: What is the units digit of (17^3)^41973^{3^2}? A. 0 B. 2 C. 4 D. 6 E. 8
Quote: 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 1113=2, which gives the final answer that the units digit of (17^3)^41973^{3^2} is 2.
Answer B. I agree with all the steps above but i have this doubt in this unit digit of 1st number is 1. Unit digit of second number is 3 13 would it not mean unit digit is 8? for eg if we have 551 853 ..Unit digit will be 2 but what if it is 551453 .unit digit will be 8? so how do we know which is the larger number of these, although it says XY? Please let me know,. That's a little trap there. Notice that the second number is much larger then the first one, thus their difference will be negative, something like 1113=2, which gives the final answer that the units digit of \((17^3)^41973^{3^2}\) is 2. So, as you can see the cases like 551453 (larger number minus smaller number) are not possible. Why is second number much larger then the first one? Consider this, even if we had \((100^3)^4\) (instead of \((17^3)^4\)) and \(1000^{(3^2)}\) (instead of \(1973^{(3^2)}\)) > \((100^3)^4=100^{12}=10^{24}\) and \(1000^{(3^2)}=1,000^9=10^{27}\). Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions
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23 Apr 2012, 11:50
Bunuel wrote: 4^{\sqrt{y} For Question #3. I'm confused why 4 squareroot Y = (2 squareroot Y)^2 instead of (2 ^2 squareroot Y). Not sure I understand the properties correctly...



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23 Apr 2012, 13:21
KGG88 wrote: Bunuel wrote: 4^{\sqrt{y} For Question #3. I'm confused why 4 squareroot Y = (2 squareroot Y)^2 instead of (2 ^2 squareroot Y). Not sure I understand the properties correctly... \(4^{\sqrt{y}}=(2^2)^{\sqrt{y}}=2^{2*\sqrt{y}}=(2^{\sqrt{y}})^2\). Check this for properties: toughandtrickyexponentsandrootsquestions125956.html#p1027888Hope it helps.
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25 Jun 2012, 09:59
kuttingchai wrote: Bunuel wrote: 2. What is the units digit of \((17^3)^41973^{3^2}\)? 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 1113=2, which gives the final answer that the units digit of \((17^3)^41973^{3^2}\) is 2.
Answer B. Hey Bunuel, I understood the logic behind finding the unit places, but how can u determine if the reminder is 2 or 8 in this example (17^12)  (1973^9) = unit place is 2 (agreed i actually did the calculation using calculator ) because : (17^12) unit place = 1 (1973^9) unit place = 3 but if we have just  (7^12)  (3^9) or (3^9)  (7^12) we have unit place as "8" 3^9 = 19683  unit place will still be 2 7^12 = 13841287201  unit place will still be 1 so my question is how will u determine if the answer is 2/8? clearly we have 2 different answers?? am i missing anything?? The units digit of \(17^{12}1973^{9}\) is 2 and not 8 since \(1973^{9}\) is much larger number than \(17^{12}\), thus their difference will be negative, something like 1113=2. If we had something like 2113 (if the first number were greater than the second one), then the units digit of their difference would be 8. Hope i's clear.
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Updated on: 15 Oct 2012, 01:35
Bunuel wrote: SOLUTIONS:
1. What is the value of \(\sqrt{25+10\sqrt{6}}+\sqrt{2510\sqrt{6}}\)? 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 unsquare 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 \((xy)^2=x^22xy+y^2\)).
So we get: \((\sqrt{25+10\sqrt{6}}+\sqrt{2510\sqrt{6}})^2=(\sqrt{25+10\sqrt{6}})^2+2(\sqrt{25+10\sqrt{6}})(\sqrt{2510\sqrt{6}})+(\sqrt{2510\sqrt{6}})^2=\) \(=(25+10\sqrt{6})+2(\sqrt{25+10\sqrt{6}})(\sqrt{2510\sqrt{6}})+(2510\sqrt{6})\).
Note that sum of the first and the third terms simplifies to \((25+10\sqrt{6})+(2510\sqrt{6})=50\), so we have \(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{2510\sqrt{6}})\) > \(50+2(\sqrt{25+10\sqrt{6}})(\sqrt{2510\sqrt{6}})=50+2\sqrt{(25+10\sqrt{6})(2510\sqrt{6})}\).
Also must know for the GMAT: \((x+y)(xy)=x^2y^2\), thus \(50+2\sqrt{(25+10\sqrt{6})(2510\sqrt{6})}=50+2\sqrt{25^2(10\sqrt{6})^2)}=50+2\sqrt{625600}=50+2\sqrt{25}=60\).
Recall that we should unsquare this value to get the right the answer: \(\sqrt{60}=2\sqrt{15}\).
Answer: C. Another way to do it, using the same formulas: \(\sqrt{25+10\sqrt{6}}+\sqrt{2510\sqrt{6}}=\sqrt{5(5+2\sqrt{6})}+\sqrt{5(52\sqrt{6})}=\sqrt{5}\sqrt{5+2\sqrt{6}}+\sqrt{5}\sqrt{52\sqrt{6}}=\) \(=\sqrt{5}(\sqrt{5+2\sqrt{6}}+\sqrt{52\sqrt{6}})=\sqrt{5}(\sqrt{(\sqrt{3}+\sqrt{2})^2}+\sqrt{(\sqrt{3}\sqrt{2})^2})=\sqrt{5}(\sqrt{3}+\sqrt{2}+\sqrt{3}\sqrt{2})=\sqrt{5}(2\sqrt{3})=2\sqrt{15}\)
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Originally posted by EvaJager on 29 Jun 2012, 11:34.
Last edited by EvaJager on 15 Oct 2012, 01:35, edited 1 time in total.



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15 Oct 2012, 00:54
Bunuel wrote: 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: \(x^2=6+x\) > \((x+2)(x3)=0\) > \(x=2\) or \(x=3\), but since \(x>0\) then: \(x=3\).
Answer: B. Hi Bunuel  All sols are crystal clear except this one logic mentioned here. I'm not able to understand the below part. Could you please elaborate? \(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}\), 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 \(x=\sqrt{6+x}\). Cheers



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15 Oct 2012, 03:09
Jp27 wrote: Bunuel wrote: 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: \(x^2=6+x\) > \((x+2)(x3)=0\) > \(x=2\) or \(x=3\), but since \(x>0\) then: \(x=3\).
Answer: B. Hi Bunuel  All sols are crystal clear except this one logic mentioned here. I'm not able to understand the below part. Could you please elaborate? \(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}\), 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 \(x=\sqrt{6+x}\). Cheers Given: \(x=\sqrt{6+({\sqrt{6+\sqrt{6+\sqrt{6+...})}}}}\). Consier the expression in brackets: \({(\sqrt{6+\sqrt{6+\sqrt{6+...}})\). It's the same as the right hand side of the initial expression, thus it also equals to \(x\). When replaced we'll have: \(x=\sqrt{6+x}\). Hope it's clear.
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Re: NEW!!! Tough and tricky exponents and roots questions
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15 Oct 2012, 03:09



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