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16 Jul 2014, 12:07
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B
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Difficulty:
45%
(medium)
Question Stats:
70% (02:04) correct
30%
(02:38)
wrong
based on 683
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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
A. -18
B. -17
C. 1
D. 17
E. 18
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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. \(18\)
Given that \((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}\), we can rewrite this as:
Next, we cross-multiply:
Now, we factorize:
To find the value of \(y\), we equate the powers of 2 on both sides:
To find the value of \(x\), we equate the powers of 7 on both sides:
Finally, we calculate \(y-x=1-18=-17\).
Answer: B
Try NEW Exponents and Roots DS question.
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. \(18\)
Given that \((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{28})^{18y}\), we can rewrite this as:
\(\frac{1}{7^x}*\frac{1}{8^{12}}=\frac{1}{28^{18y}}\).
Next, we cross-multiply:
\(28^{18y}=7^x*8^{12}\).
Now, we factorize:
\(2^{36y}*7^{18y}=7^x*2^{36}\).
To find the value of \(y\), we equate the powers of 2 on both sides:
\(36y=36\), hence \(y=1\).
To find the value of \(x\), we equate the powers of 7 on both sides:
\(18*1=x\), hence \(x=18\);
Finally, we calculate \(y-x=1-18=-17\).
Answer: B
Try NEW Exponents and Roots DS question.
General Discussion
16 Jul 2014, 17:53
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Bunuel
\((\frac{1}{7})^x*(\frac{1}{8})^{12}=(\frac{1}{7})^{18y}(\frac{1}{4})^{18y}\)
For Now, exclude \((\frac{1}{7})\) to find \(Y\)
\((\frac{1}{8})^{12}=(\frac{1}{4})^{18y}\)
\((\frac{1}{2})^{36}=(\frac{1}{2})^{36y}\)
\(36y=36\) --> \(y=1\)
\((\frac{1}{7})^x=(\frac{1}{7})^{18y}\)
\((\frac{1}{7})^x=(\frac{1}{7})^{18(1)}\)
\(x=18\)
\(y-x=1-18=-17\)
Answer is B
