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x=y^3 and y>1

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Carcass
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x=y^3 and y>1 [#permalink] New post   02 Jun 2020, 02:06
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\(x=y^3\) and \(y>1\)


Quantity A
Quantity B
\(x^y\)
\(y^x\)



A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
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D
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Re: x=y^3 and y>1 [#permalink] New post   02 Jun 2020, 03:22
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This question picking numbers does not work very well

Better to use a general approach

We do know that \(x=y^3\)

Substitute in the two quantities

we do have

QA \(y^{{y{^3}}\)

QB \(y^{y+y+y}\)

We do have the same base so is the same to write

\(y^3>3y\)

\(y^3-3y>0\)

\(y(y^2-3)>0\)

Now

\(y>0\) which is false because we do know that \(y>1
\)

So we have to consider only \(y^2>3\)

Take the square root we do have that

\(y>1.7\)

We do know that \(y>1\)

However, we do have also \(y>1.7\)

Therefore y can be 1.1 or 2. Below 1.7 or greater than 1.7

The answer is D
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Re: x=y^3 and y>1 [#permalink] New post   02 Jun 2020, 02:07
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Post A Detailed Correct Solution For The Above Questions And Get A Kudos.
Question From Our New Project: GRE Quant Challenge Questions Daily - NEW EDITION!
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Re: x=y^3 and y>1 [#permalink] New post   02 Jun 2020, 02:40
Any explanations, I'm confused a little but.
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Re: x=y^3 and y>1 [#permalink] New post   07 Jun 2020, 11:09
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I picked y=2
x=y^3
x=2^3
x=8

Statement A: x^y
= 8^2
=64

Statement B: y^x
=2^8
=256

if y=1.5
x=y^3
x=3.375

Statement A: x^y
3.375^1.5

Now with decimal it is getting complicated so I dont know what to do :(
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Re: x=y^3 and y>1 [#permalink] New post   08 Jun 2020, 00:55
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I would do the following to simplify the problem:

As x = y^3

Put y^3 in QTY A:

=> y^3y(QTY A) ?? y^x (QTY B)

If we Equate

=> 3y ?? x

Now, test with Numbers:

Put y = 2

QTY A => 6
QTY B => 8

Put y = 1.1

QTY A => 3.3
QTY B => 1.331

Hence, D i.e Relationship Can't be determined.
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Re: x=y^3 and y>1 [#permalink] New post   29 Aug 2021, 21:50
When i solved this using generic method

QA : y ^ 3y

QB : y ^ y ^ 3
Therefore since bases are same you can compare , you get

QA : 3y and QB : y ^ 3

And if u substitute values ans is D
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Re: x=y^3 and y>1 [#permalink] New post   02 Sep 2021, 18:25
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Picking numbers is a good strategy for this.

Another way you could view this is:

\(y^{3y} = y^{y+y+y}\)

\(y^{y^3} = y^{y*y*y}\)

It's clear that if y = 2, then \(y^{y^3}\) is greater, but were told that \(y>1\). What about the numbers less than 2?

Let's try 1.1 for example (and ignore subbing in the value for the base for clarity):

\( y^{y+y+y} = y^{1.1+1.1+1.1} = y^{3.3}\)

\(y^{y*y*y} = y^{1.1*1.1*1.1} = y^{1.331}\)

So this in this case we have that \(y^{3y}\) is greater.

So the answer is D
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Re: x=y^3 and y>1 [#permalink] New post   27 May 2023, 20:49
Carcass how does x^y become y^y^3? could you pls give the steps
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Re: x=y^3 and y>1 [#permalink] New post   27 May 2023, 23:04
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Because of a simple substitution of our equation into QA. I assume this is what you meant

x=y^3

And because in the equation above x is the same to say y^3

in QA we do have \(x^y\) and \(x=y^3\)

so our x will become \(y^3\)\ raised to y

the result could be written as follow

\(y^{y^3} = y^{y*y*y}\)

See more theory of exponents here https://gre.myprepclub.com/forum/gre-ma ... 24948.html
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Re: x=y^3 and y>1 [#permalink] New post   03 Mar 2025, 10:35
Carcass wrote:
This question picking numbers does not work very well

Better to use a general approach

We do know that \(x=y^3\)

Substitute in the two quantities

we do have

QA \(y^y{^3\)

QB \(y^{y+y+y}\)

We do have the same base so is the same to write

\(y^3>3y\)

\(y^3-3y>0\)

\(y(y^2-3)>0\)

Now

\(y>0\) which is false because we do know that \(y>1
\)

So we have to consider only \(y^2>3\)

Take the square root we do have that

\(y>1.7\)

We do know that \(y>1\)

However, we do have also \(y>1.7\)

Therefore y can be 1.1 or 2. Below 1.7 or greater than 1.7

The answer is D


Even when we substitute x^y with y^3 we get y^3y at least this is what I got, pls explain why I am getting different when substituting
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Re: x=y^3 and y>1 [#permalink] New post   04 Mar 2025, 05:15
Carcass In this perticular question I understood till we solved exponent 3y vs y^3 after that I am not able to understand the solutions given even by u, pls help
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Re: x=y^3 and y>1 [#permalink] New post   04 Mar 2025, 08:26
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Sir

Could please point out the steps you did not gut specifically.

So I would be able to explain.

Thanks
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Re: x=y^3 and y>1 [#permalink] New post   05 Mar 2025, 04:49
Carcass wrote:
Sir

Could please point out the steps you did not gut specifically.

So I would be able to explain.

Thanks


The step where we equated the base and then equated the exponent 3y vs y^3
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x=y^3 and y>1 [#permalink] New post   05 Mar 2025, 07:45
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That is a rule of exponent sir https://gre.myprepclub.com/forum/gre-ma ... 24948.html

Suppose you have \(a^3= a^{x+y}\)

You can equate \(3=x+y\)

Because the base is the same
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Re: x=y^3 and y>1 [#permalink] New post   05 Mar 2025, 23:13
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Carcass wrote:
That is a rule of exponent sir https://gre.myprepclub.com/forum/gre-ma ... 24948.html

Suppose you have \(a^3= a^{x+y}\)

You can equate \(3=x+y\)

Because the base is the same


Yes that I got
what i meant is from the equating exponent step I am not able to solve further, the equal base so equating exponent step I understood.
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Re: x=y^3 and y>1 [#permalink] New post   05 Mar 2025, 23:58
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Let me know if you need further explanations or help for your preparation for the GRE.

I would be happy to step in once again sir
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Re: x=y^3 and y>1 [#permalink]
05 Mar 2025, 23:58
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