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(x^y) * (x^2)=x^6

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Carcass
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(x^y) * (x^2)=x^6 [#permalink] New post   29 Jun 2020, 08:33
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\((x^y) \times (x^2)=x^6\) and \((2z)^{-2} = \frac{1}{36}\)

Quantity A
Quantity B
\(\frac{1}{y}\)
\(\frac{1}{z}\)



A)The quantity in Column A is greater.
B)The quantity in Column 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) * (x^2)=x^6 [#permalink] New post   30 Jun 2020, 13:14
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Carcass wrote:
\((x^y) \times (x^2)=x^6\) and \((2z)^{-2} = \frac{1}{36}\)

Quantity A
Quantity B
\(\frac{1}{y}\)
\(\frac{1}{z}\)




There are lots of problems with this question

Given: \((2z)^{-2} = \frac{1}{36}\)

Rewrite as: \(\frac{1}{(2z)^2} = \frac{1}{36}\)

Simplify: \(\frac{1}{4z^2} = \frac{1}{36}\)

From this we can conclude that: \(4z^2 = 36\)

Divide both sides by 4 to get: \(z^2 = 9\)

So, z = 3 OR z = -3


Given: \((x^y) \times (x^2)=x^6\)
One possible solution here is: x = 5 and y = 4, since (5^4)(5^2) = 5^6
Another possible solution is: x = 0 and y = 99, since (0^99)(0^2) = 0^6
Another possible solution is: x = 1 and y = -7, since (1^77)(1^2) = 1^6
.
.
.etc

Since y can have infinitely many values, the correct answer here is D

Cheers,
Brent
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Re: (x^y) * (x^2)=x^6 [#permalink] New post   29 Jun 2020, 08:33
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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) * (x^2)=x^6 [#permalink] New post   29 Jun 2020, 10:26
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So first we know that (x^y) * (x^2) = x^6
When we multiply two numbers that are expressed as x raised to some power, the result is x raised to the power of the sum of the two powers
So in our case 6 = y+2 which means y=4

Next:
(2z)^−2= 1/36

(2z)^-2 = 1/(2z)^2 = 1/(4z^2)

Now since 1/(4z^2) = 1/36 , we calculate z = 3

Lastly, we just compare 1/3 to 1/4 resulting in B as the right answer
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Re: (x^y) * (x^2)=x^6 [#permalink] New post   29 Jun 2020, 10:29
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\((x^y) \times (x^2)=x^6\)

\(x^{y+2}=x^6\)

\(y=4\)

\((2z)^{-2} = \frac{1}{36}\)

\(\frac{1}{(2z)^2}=\frac{1}{36 }\)

\(\frac{1}{2^2 \times z^2}=6^2\)

Basically

\(4z^2=36\)

\(z^2=9\)

\(z=3\)

So y=4 and z=3

A is \(\frac{1}{4}\) and B is \(\frac{1}{3}\)

B is greater
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Re: (x^y) * (x^2)=x^6 [#permalink] New post   30 Jun 2020, 08:45
Carcass wrote:
\((x^y) \times (x^2)=x^6\)

\(x^{y+2}=x^6\)

\(y=4\)

\((2z)^{-2} = \frac{1}{36}\)

\(\frac{1}{(2z)^2}=\frac{1}{36 }\)

\(\frac{1}{2^2 \times z^2}=6^2\)

Basically

\(4z^2=36\)

\(z^2=9\)

\(z=3\)

So y=4 and z=3

A is \(\frac{1}{4}\) and B is \(\frac{1}{3}\)

B is greater


Could help explain why can't z also be -3?
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Re: (x^y) * (x^2)=x^6 [#permalink] New post   30 Jun 2020, 10:24
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Attachment:
shot122.jpg
shot122.jpg [ 257.46 KiB | Viewed 6860 times ]


We have created the Quant math book for this purpose.

https://gre.myprepclub.com/forum/greprepcl ... -2609.html

Use it :wink:
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Re: (x^y) * (x^2)=x^6 [#permalink] New post   30 Jun 2020, 13:03
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So, Z can be +3 or -3 right?
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Re: (x^y) * (x^2)=x^6 [#permalink] New post   Updated on: 30 Jun 2020, 13:15
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Carcass wrote:
Attachment:
shot122.jpg


We have created the Quant math book for this purpose.

https://gre.myprepclub.com/forum/greprepcl ... -2609.html

Use it :wink:



I didn't know about this. Thanks a lot, it's very good.
But I am even more confused now, because as it says in this Quant book:

Carcass wrote:
This post is a part of [GRE MATH BOOK]


Exponential Equations:
When solving equations with even exponents, we must consider both positive and negative possibilities for the solutions.

For instance \(a^2=25\), the two possible solutions are \(5\) and \(-5\).


so in our example, z^2 = 9 ; z = 3 or - 3 ? right?
can you help shed some light here?

Originally posted by tpaillier on 30 Jun 2020, 13:14.
Last edited by tpaillier on 30 Jun 2020, 13:15, edited 1 time in total.
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Re: (x^y) * (x^2)=x^6 [#permalink] New post   30 Jun 2020, 14:20
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Clear the way guys.

IF the variable is x^2= some value for instance x^2=25

x is equal to 5 AND minus 5

And the book is right. After all this is a universal math law.

Back to the question, I checked out the book from which I took it and YEs the book is wrong and @greenlighttest prep is RIGHT (as almost always) and the GRE Quant book I made is Correct . However, the book from the question comes from is WRONGG

\(z^2= 9\) ---------> z=+3 or -3

So in one case the answer is B and in the other case the answer is A

So the definite answer is D

Many Thanks
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