If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be : PS Archive

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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

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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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

09 Dec 2017, 16:52

Expert Reply

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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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

09 Dec 2017, 23:42

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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 genxer123

No, 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]

10 Dec 2017, 14:46

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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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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

10 Dec 2017, 20:36

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Expert Reply

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 counter-intuitive, 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 here

And 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]

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]

11 Dec 2017, 20:24

Expert Reply

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| ?

P

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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

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

L

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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

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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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

12 Dec 2017, 21:41

Expert Reply

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.

P

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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

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

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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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Re: If a^3 = 36, b^4 = 81, and c^5 = 125, which of the following must be [#permalink]

07 Jun 2019, 07:04