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.

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.

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.

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.

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

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

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 \(36y = 36\)