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If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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07 Dec 2017, 21:22
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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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07 Dec 2017, 23:50
Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b Since \(a^3 = 36\), a is positive. The value of a is in the range 3 < a < 4 because \(a^3 = 36\), which is between \(27(3^3) and 64(4^3)\) Since \(b^4 = 81\), b could be both positive or negative. The value of b is 3. Since \(c^5 = 125\), c is positive. The value of c is in the range 2 < c < 3 because \(c^5 = 125\), which is between \(32(2^5) and 243(3^5)\) Hence, a>b>c(Option A) must be true.
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If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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Updated on: 09 Dec 2017, 17:17
Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b This question is a lot harder than it looks, IMO. I'd bet many will get lucky when choosing A. If \(n^{odd} = positive\) number, \(n\) is positive Approximate value of \(a\)\(a^3 = 36\), and \(36\) is positive \(a\) must be positive The closest integer for \(a\) is \(3\), because \(3^3 = 27\), so \(a\) must be a little bigger than \(3\) \(a = 3+\) Value of \(b\)\(b^4 = 81\) \(b = 3\) or \(b = 3\) Approximate value of \(c\)\(c^5 = 125\) \(125\) is positive. \(c\) must be positive The two integers closest to \(c\) are \(2\) and \(3\): \(2^5=32\) and \(3^5=243\), where \((2^5=32) < (c^5=125) < (3^5=243)\) Compare \(c\) to positive \(b\) \(c\) cannot be \(3\) \(c^5=125 \neq{243}\) (not even close) \(c \neq{3} \neq{+b}\) \(c = 2+\) Is \(b\) positive or negative?If \(b = +3\) \(a > b > c\) (Answer A) If \(b = 3\) \(a > c > b\) (Answer B) Is \(b\) positive or negative? A quandary, maybe. The principal root rule is no help.* Negative numbers raised to even powers: bracketsI think the key is the "bracket" rule for negative numbers raised to even powers. Think of "negative four, squared." IF \(4\) has brackets, then: \((4)^2 = 16\) That means: \((4)(4) = 16\) BUT IF \(4\) has no brackets, then: \(4^2 = 16\) That means: \( [(4)(4)] = 16\) Expanding: \(4^2 = ?\) By PEMDAS order of operations, we take exponents first. So first we raise \(4\) to the second power: \(4^2 = 16\) Then we take the negative of the result: \(16\) \(4^2 = 16\) \(b^4 = 81\) IF \(b\) were negative, in order to equal positive 81, \(b\) would REQUIRE brackets. Per above: \((3)^4 = 81\) BUT \(3^4 = 81\) There are no brackets around \(b\). So \(b\) in \(b^4 = 81\) MUST be positive \(b = 3\), and \(a = 3+\) \(c = 2+\) \(a > b > c\) ANSWER A Please correct me if I am mistaken. I see no other way to choose A over B. The absolute value of \(b = 3\), but I do not see how absolute value tells us the sign of \(b\). * That is IF we are GIVEN \(\sqrt[4]{b^4}\) , the answer \(b\), is positive. The radical sign itself tells us to take the principal fourth root, the positive one.
But we are given an exponent, not a radical sign, such that \(b^4 = 81\) has two solutions: \(3\) and \(3\). In other words, given \(b^4 = 81\), there is no radical sign to indicate "Take the principal (positive) fourth root." Without such direction, we still do not know if \(b\) is positive or negative.
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Originally posted by generis on 09 Dec 2017, 16:33.
Last edited by generis on 09 Dec 2017, 17:17, edited 4 times in total.



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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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09 Dec 2017, 16:39
Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b a^3=36 Hence, a must be positive since when x is raised to an odd power, it retains it's original sign. Also, 3<a<4. b^4=81 Hence, b could be positive or negative. b could be 3 or +3. So, b<a. c^5 = 125 Similar to the case of a above, c has to be positive. Also, 2<c<3. Hence, c<b<a. Answer is A.
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If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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09 Dec 2017, 16:52
Madnov2017 wrote: Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b a^3=36 Hence, a must be positive since when x is raised to an odd power, it retains it's original sign. Also, 3<a<4. b^4=81 Hence, b could be positive or negative. b could be 3 or +3. So, b<a. c^5 = 125 Similar to the case of a above, c has to be positive. Also, 2<c<3. Hence, c<b<a. Answer is A. Madnov2017 and pushpitkc : If a and c must be positive, and b can be positive or negative, how did you decide that b must be positive? No disrespect, I promise: Neither of you mentions that b must be positive for Answer A to be correct. Neither do you explain why b is positive. It seems as if you kinda skipped over the issue? Yes, no matter what, a > b. a = 3+, and b = 3 OR 3 a > b The question seems to me to be: Is b > c, or is c > b? c = 2+ (and cannot = 3) b = 3 OR 3 If b = 3, c > b, and Answer B is correct. If b = +3, b > c, and Answer A is correct. How to choose the sign of b seems to me to be the only way to know which is greater, b or c Am I missing something?
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If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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09 Dec 2017, 23:42
genxer123 wrote: Madnov2017 wrote: Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b a^3=36 Hence, a must be positive since when x is raised to an odd power, it retains it's original sign. Also, 3<a<4. b^4=81 Hence, b could be positive or negative. b could be 3 or +3. So, b<a. c^5 = 125 Similar to the case of a above, c has to be positive. Also, 2<c<3. Hence, c<b<a. Answer is A. Madnov2017 and pushpitkc : If a and c must be positive, and b can be positive or negative, how did you decide that b must be positive? No disrespect, I promise: Neither of you mentions that b must be positive for Answer A to be correct. Neither do you explain why b is positive. It seems as if you kinda skipped over the issue? Yes, no matter what, a > b. a = 3+, and b = 3 OR 3 a > b The question seems to me to be: Is b > c, or is c > b? c = 2+ (and cannot = 3) b = 3 OR 3 If b = 3, c > b, and Answer B is correct. If b = +3, b > c, and Answer A is correct. How to choose the sign of b seems to me to be the only way to know which is greater, b or c Am I missing something? Hi genxer123No, I believe that you maybe correct. Because in my solution, I have assumed that b is positive. However, if b is negative then Option B(a>c>b) could also be true. Hello Bunuel Can you please help us with a solution for this problem. Thanks in advance.
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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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10 Dec 2017, 14:46
genxer123 wrote: Madnov2017 wrote: Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b a^3=36 Hence, a must be positive since when x is raised to an odd power, it retains it's original sign. Also, 3<a<4. b^4=81 Hence, b could be positive or negative. b could be 3 or +3. So, b<a. c^5 = 125 Similar to the case of a above, c has to be positive. Also, 2<c<3. Hence, c<b<a. Answer is A. Madnov2017 and pushpitkc : If a and c must be positive, and b can be positive or negative, how did you decide that b must be positive? No disrespect, I promise: Neither of you mentions that b must be positive for Answer A to be correct. Neither do you explain why b is positive. It seems as if you kinda skipped over the issue? Yes, no matter what, a > b. a = 3+, and b = 3 OR 3 a > b The question seems to me to be: Is b > c, or is c > b? c = 2+ (and cannot = 3) b = 3 OR 3 If b = 3, c > b, and Answer B is correct. If b = +3, b > c, and Answer A is correct. How to choose the sign of b seems to me to be the only way to know which is greater, b or c Am I missing something? Apologies. It seems like I did skip over this issue. I can't think of a way to find out the sign of b. I see that genxer123 makes use of some parenthesis rule to figure out the sign of b. However, I am not sure such a rule exists and don't think it works here. b can take a value of 3 or 3 and for b^4=81, b has to be within parenthesis, right? If we take '''' outside of the parenthesis, then the original equation b^4=81 fails, isn't it? I hope I am not missing something very obvious. Don't want to sound too dumb. :D
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If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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10 Dec 2017, 20:36
Madnov2017 wrote: Apologies. It seems like I did skip over this issue. I can't think of a way to find out the sign of b. I see that genxer123 makes use of some parenthesis rule to figure out the sign of b. However, I am not sure such a rule exists and don't think it works here. b can take a value of 3 or 3 and for b^4=81, b has to be within parenthesis, right? If we take '''' outside of the parenthesis, then the original equation b^4=81 fails, isn't it? I hope I am not missing something very obvious. Don't want to sound too dumb. :D Dumb you most certainly neither appear to be, sound, nor are. Quote: I am not sure such a rule exists It does. It is not obvious. That is exactly why Bunuel has it skulking around here. Perhaps it is not a rule. Perhaps it is a "worldwide convention"? An "arithmetic principle"? I don't know what to call it. I just know it exists. It can seem counterintuitive, but: \(3^4 = 81\) \((3)^4 = 81\) If there is a negative number raised to an even power: AND no parentheses? The result is negative. Quote: [If such a rule exists, I] don't think it works here. b can take a value of 3 or 3 and for b^4=81, b has to be within parenthesis, right? No. For negative \(b^4\) to equal positive \(81\), the negative \(b^4\) must be in parentheses. Positive \(b^4\) does NOT have to be in parentheses. That is precisely why the rule DOES work. Only positive \(b^4\) with no parentheses can = positive 81. So \(b\) must be positive. THUS \(3^4 = 81\) \((3)^4 = 81\) (allowed, but not conventional) \(3^4 =  81\) \((3)^4 = 81\) If we had seen parentheses, \((b)^4 = 81\), we would have been stuck without a way to discern b's sign. Both positive and negative 3 CAN go in brackets to get a positive result. Negative 3 must go in brackets to get a positive result. Both are possible, one is required, and there would have been no way to tell the difference. But instead we saw \(b^4=81\) Substitute \(3\) and \(3\) for \(b\): \(3^4 = 81\) \(3^4 = 81\) So we DO have a way to discern \(b\)'s sign: No brackets? Positive result? \(b\) must be positive Quote: I can't think of a way to find out the sign of b. I can. Use the "negative number raised to even power without brackets" rule: no brackets and a positive result? The number b is positive. And that is the only reason I can find for "b must be positive." Thanks, pushpitkc and Madnov2017 , for responding. And kudos. I appreciate it. This question's layers are brilliant. Wish I could give its writer extra kudos. Below are three sources who address the issue. purplemath guy says here: Quote: Simplify \((–3)^2\)The square means "multiplied against itself, with two copies of the base". This means that I'll have two "minus" signs, which I can cancel:
\((–3)^2 = (–3)(–3) = (+3)(+3) = 9\) Pay careful attention and note the difference between the above exercise and the following:
Simplify \(–3^2\) \(–3^2 = –(3)(3) = –1(3)(3) = (–1)(9) = –9\) In the second exercise, the square (the "to the power 2") was only on the 3; it was not on the minus sign. Those parentheses in the first exercise make all the difference in the world! Be careful with them. . . Quote: \(−3^2\) does not mean "the square of negative three." The exponent takes priority over the negative: it means "the negative of \(3^2\)." See hereAnd finally, from the Monterrey Institute: Quote: That leaves us with the term \(3^4.\) This example is a little trickier because there is a negative sign in there. One of the rules of exponential notation is that the exponent relates only to the value immediately to its left. So, 3^4 does not mean 3 • 3 • 3 • 3. It means “the opposite of 3^4,” or — (3 • 3 • 3 • 3). If we wanted the base to be 3, we’d have to use parentheses in the notation: \((3)^4.\) Why so picky? Well, do the math: 3^4 = – (3 • 3 • 3 • 3) = 81 (3)^4 = 3 • 3 • 3 • 3 = 81 That’s an important difference. My emphases. That material is HERE
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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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11 Dec 2017, 19:31
Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b I can think of the following logics to establish b sign. Please let me know if I am wrong in the following logic. 1. I think as a,b,c are close to an AP  Hence a >b>c. 2. b > c.. Square on both sides  Valid. c > b.. Square on both sides  Not valid Hi Bunuel and Forum Experts, Can you please help me with the below doubt. 1. In an inequality can we apply mod on both sides. For Ex  Suppose k > m .. can we write k > m.. Please advise.
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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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11 Dec 2017, 20:24



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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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11 Dec 2017, 20:32
Bunuel wrote: rahul16singh28 wrote: Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b I can think of the following logics to establish b sign. Please let me know if I am wrong in the following logic. 1. I think as a,b,c are close to an AP  Hence a >b>c. 2. b > c.. Square on both sides  Valid. c > b.. Square on both sides  Not valid Hi Bunuel and Forum Experts, Can you please help me with the below doubt. 1. In an inequality can we apply mod on both sides. For Ex  Suppose k > m .. can we write k > m.. Please advise. 2 > 3. Is 2 > 3 ? Thanks for clarifying the doubt Bunuel
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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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12 Dec 2017, 15:21
pushpitkc wrote: Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b Since \(a^3 = 36\), a is positive. The value of a is in the range 3 < a < 4 because \(a^3 = 36\), which is between \(27(3^3) and 64(4^3)\) Since \(b^4 = 81\), b could be both positive or negative. The value of b is 3. Since \(c^5 = 125\), c is positive. The value of c is in the range 2 < c < 3 because \(c^5 = 125\), which is between \(32(2^5) and 243(3^5)\) Hence, a>b>c(Option A) must be true. How could you say value of B is 3. It could be negative. In that case Option B will be valid.



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If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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12 Dec 2017, 21:41
ammuseeru wrote: pushpitkc wrote: Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b Since \(a^3 = 36\), a is positive. The value of a is in the range 3 < a < 4 because \(a^3 = 36\), which is between \(27(3^3) and 64(4^3)\) Since \(b^4 = 81\), b could be both positive or negative. The value of b is 3. Since \(c^5 = 125\), c is positive. The value of c is in the range 2 < c < 3 because \(c^5 = 125\), which is between \(32(2^5) and 243(3^5)\) Hence, a>b>c(Option A) must be true. How could you say value of B is 3. It could be negative. In that case Option B will be valid. ammuseeru , maybe you didn't read the whole topic? We have been discussing exactly that.
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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]
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15 Dec 2017, 07:05
Bunuel wrote: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be true?
A. a>b>c B. a>c>b C. b>a>c D. b>c>a E. c>a>b Hi Bunuel, Can you please provide the OE for this question. Marked this day in my calendar for your solution == Message from GMAT Club Team == This is not a quality discussion. It has been retired. If you would like to discuss this question please repost it in the respective forum. Thank you! To review the GMAT Club's Forums Posting Guidelines, please follow these links: Quantitative  Verbal Please note  we may remove posts that do not follow our posting guidelines. Thank you.
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