Official 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: \(18*1=x\) --> \(x=18\);

\(y-x=1-18=-17\).

Answer: B.
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I'm a little rusty, how do you cross multiply to get everything to be a divisor of 1 to being the number alone to the exponent. That is where I struggled with this problem. Thanks.

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

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

Cross-multiply: \(28^{18y}=7^x*8^{12}\).
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Not sure I am following when you equate the exponents of 7. Would that not be 18y=x?

Thanks for the help.

Yes, but from the line just above it we established that y = 1, thus 18*1 = x.
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Hello,

How does the left hand side of this equation factor? ie the 28^18y

Sorry, a bit rusty on the the exponent rules. Thanks in advance.

\(28^{18y}\);

\((4*7)^{18y}\);

\((2^2*7)^{18y}\);

\((2^2)^{18y}*7^{18y}\);

\(2^{36y}*7^{18y}\).

Hope it's clear.
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I rewrote the equation as:
7^-x∗8^-12=28^-18y
7^-x∗2^3(-12) = 7^-18y∗ 2^2(-18y)
7^-x∗2^-36 = 7^-18y∗ 2^-36y
from here we can see that -x = -18y or x = 18y and -36=-36y or y =1.
If y is 1 then x = 18
y-x = 1-18 = -17

Bunuel
Hi Bunuel, if x & y are not integers, does it give infinitely many solutions of x & y, or does it still give finite (but different) solutions? Thanks!

It will have infinite number of solutions. For every x, there will be some y to satisfy the equation and vise-versa.
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i did this question in following way:
simplified the given equation into 7^(18y-x) = 2^36(1-y)
now as 7 and 2 are prime no.s , LHS will be equal to RHS only when when respective powers of 7 and 2 is 0
hence, i put 18y-x=0, this implies 18y=x.
and 36(1-y)=0, this implies y=1
hence y-x= 1-18= -17

is above method correct?

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Yes, it's correct.
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You can cross multiply any fraction where the dividend is 1 by just bringing up the divisor. In this question 1/28^18y is the same as 1/1 * 28^18y/1. If you cross multiply this you should get just 28^18y. The same rule applies to the other fractions in the equation. Hope this helps!

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Without doing manipulation of exponents to solve for y and x individually:

For y-x, we know that RHS = (1/7)^18y * (1/4)^18y

LHS = RHS on (1/7) where x= 18y

y-x = y-18y = -17y; since y is > 0 and integer, we would not have an answer choice where the expression = -17(2) or -17(3) since there is only 1 right answer. Therefore, -17 is definitely the right answer with -17(1).

Answer --> (B)

Please let me know where I am going wrong?
8^-12= 4^-18y

4^-24 = 4^-18y

8^(-12) does not equal to 4^(-24).


8^(-12) = (2^3)^(-12) = 2^(-36), while 4^(-24) = (2^2)^(-24) = 2^(-48).
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