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Math Expert V
Joined: 02 Sep 2009
Posts: 55729
If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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Difficulty:   45% (medium)

Question Stats: 70% (01:58) correct 30% (02:32) wrong based on 221 sessions

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If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

Kudos for a correct solution.

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Math Expert V
Joined: 02 Sep 2009
Posts: 55729
If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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SOLUTION

If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

$$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$;

Cross-multiply: $$28^{18y}=7^x*8^{12}$$;

Factorize: $$2^{36y}*7^{18y}=7^x*2^{36}$$;

Equate the powers of 2 on both sides: $$36y=36$$ --> $$y=1$$;

Equate the powers of 7 on both sides: $$18*1=x$$ --> $$x=18$$;

$$y-x=1-18=-17$$.

Try NEW Exponents and Roots DS question.
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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Bunuel wrote:

If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

Kudos for a correct solution.

$$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{7})^{18y}(\frac{1}{4})^{18y}$$

For Now, exclude $$(\frac{1}{7})$$ to find $$Y$$
$$(\frac{1}{8})^{12}=(\frac{1}{4})^{18y}$$
$$(\frac{1}{2})^{36}=(\frac{1}{2})^{36y}$$

$$36y=36$$ --> $$y=1$$

$$(\frac{1}{7})^x=(\frac{1}{7})^{18y}$$
$$(\frac{1}{7})^x=(\frac{1}{7})^{18(1)}$$
$$x=18$$

$$y-x=1-18=-17$$

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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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Focusing on denominator of $$(\frac{1}{8})^{12}$$ to adjust to $$\frac{1}{4}$$

$$8^{12} = 2^{(3*12)} = 2^{36} = 2^{(2*18)} = 4^{18}$$

$$(\frac{1}{7})^x * (\frac{1}{4})^{18} = (\frac{1}{7})^{18y} * (\frac{1}{4})^{18y}$$

Equating powers of similar terms:

x = 18y & 18 = 18y

y = 1; x = 18

y-x = 1-18 = -17

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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

Sol: The given expression can be re-written as $$(\frac{1}{7})^x * (\frac{1}{2^3})^{12}= (\frac{1}{7})^{18y} *(\frac{1}{2^2})^{18y}$$

Equating powers we get on both sides

x=18y
and 36=36y or y =1
x=18

y-x=-17.

Ans is B
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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Bunuel wrote:

If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

Kudos for a correct solution.

$$\frac{1}{7^x}*\frac{1}{8^{12}}=\frac{1}{7^{18y}}*\frac{1}{2^{36y}}$$
$$\frac{1}{7^x}*\frac{1}{2^{36}}=\frac{1}{7^{18y}}*\frac{1}{2^{36y}}$$
y = 1
x=18
y-x = -17
Math Expert V
Joined: 02 Sep 2009
Posts: 55729
Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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SOLUTION

If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

$$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$;

Cross-multiply: $$28^{18y}=7^x*8^{12}$$;

Factorize: $$2^{36y}*7^{18y}=7^x*2^{36}$$;

Equate the powers of 2 on both sides: $$36y=36$$ --> $$y=1$$;

Equate the powers of 7 on both sides: $$18*1=x$$ --> $$x=18$$;

$$y-x=1-18=-17$$.

Kudos points given to correct solutions.

Try NEW Exponents and Roots DS question.
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If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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Bunuel wrote:

If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

Kudos for a correct solution.

$$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$

Or, $$(\frac{1}{7})^x*(\frac{1}{2^3})^{12}=(\frac{1}{2^2*7})^{18y}$$

Or, $$(\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})$$

Now, $$36y = 36$$

So, $$y = 1$$ and $$x = 18$$

Or, $$y - x$$ = $$1 - 18$$

So, Answer will be (B) $$- 17$$
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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Bunuel wrote:
SOLUTION

If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

$$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$;

Cross-multiply: $$28^{18y}=7^x*8^{12}$$;

Factorize: $$2^{36y}*7^{18y}=7^x*2^{36}$$;

Equate the powers of 2 on both sides: $$36y=36$$ --> $$y=1$$;

Equate the powers of 7 on both sides: $$18*1=x$$ --> $$x=18$$;

$$y-x=1-18=-17$$.

Try NEW Exponents and Roots DS question.

Hi pushpitkc

from here $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$;

i get $$(\frac{1}{56})^{12+x}=(\frac{1}{28})^{18y}$$; after this i get lost, can you please explain how Bunuel arrived at correct answer? thank you Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3372
Location: India
GPA: 3.12
Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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dave13 wrote:
Bunuel wrote:
SOLUTION

If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

$$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$;

Cross-multiply: $$28^{18y}=7^x*8^{12}$$;

Factorize: $$2^{36y}*7^{18y}=7^x*2^{36}$$;

Equate the powers of 2 on both sides: $$36y=36$$ --> $$y=1$$;

Equate the powers of 7 on both sides: $$18*1=x$$ --> $$x=18$$;

$$y-x=1-18=-17$$.

Try NEW Exponents and Roots DS question.

Hi pushpitkc

from here $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$;

i get $$(\frac{1}{56})^{12+x}=(\frac{1}{28})^{18y}$$; after this i get lost, can you please explain how Bunuel arrived at correct answer? thank you Hi dave13

The rule for exponents is $$a^x * a^y = a^{x+y}$$
However, if we have different bases, we cannot add the exponents $$a^x * b^y$$ is not equal to $$ab^{x+y}$$
As a result, what you have done is wrong. Bunuel has solved this problem in the easiest possible way.

Check his solution here. If you have any doubts, please share it. I will try to help you _________________
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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Abhishek009 wrote:
Bunuel wrote:

If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

Kudos for a correct solution.

$$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$

Or, $$(\frac{1}{7})^x*(\frac{1}{2^3})^{12}=(\frac{1}{2^2*7})^{18y}$$

Or, $$(\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})$$

Now, $$36y = 36$$

So, $$y = 1$$ and $$x = 18$$

Or, $$y - x$$ = $$1 - 18$$

So, Answer will be (B) $$- 17$$

hey pushpitkc thanks for the hint, but i stil dont get how from this $$(\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})$$

we get this $$36y = 36$$ Senior PS Moderator V
Joined: 26 Feb 2016
Posts: 3372
Location: India
GPA: 3.12
If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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dave13 wrote:
Abhishek009 wrote:
Bunuel wrote:

If $$x$$ and $$y$$ are positive integers and $$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$, then what is the value of $$y-x$$?

A. -18
B. -17
C. 1
D. 17
E. 18

Kudos for a correct solution.

$$(\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}$$

Or, $$(\frac{1}{7})^x*(\frac{1}{2^3})^{12}=(\frac{1}{2^2*7})^{18y}$$

Or, $$(\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})$$

Now, $$36y = 36$$

So, $$y = 1$$ and $$x = 18$$

Or, $$y - x$$ = $$1 - 18$$

So, Answer will be (B) $$- 17$$

hey pushpitkc thanks for the hint, but i stil dont get how from this $$(\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})$$

we get this $$36y = 36$$ Hi dave13

We know that $$1^{anything} = 1$$

$$(\frac{1}{7})^x*(\frac{1}{2})^{36}=(\frac{1}{2^{36y}*7^{18y}})$$ --> $$\frac{1}{7^x}*\frac{1}{2^{36}}=(\frac{1}{2^{36y}*7^{18y}})$$

Cross-multiplying, we will get $$2^{36y}*7^{18y} = 7^x*2^{36}$$

If we have $$a^x*b^y = a^w*b^z$$, x = w and y = z

Applying this learning to the above equation, we will get 36y = 36 | 18y = x

$$y = \frac{36}{36} = 1$$
$$18y = x$$ -> $$x = 18$$ (because y = 1)

Therefore, the value of the expression y-x is 1-18 = -17(Option B)

Hope this helps you!
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Re: If x and y are positive integers and (1/7)^x*(1/8)^12=(1/28)  [#permalink]

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