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16 Sep 2014, 00:34
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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. \(18\)
16 Sep 2014, 00:34
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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:
\(\frac{1}{7^x}*\frac{1}{8^{12}}=\frac{1}{28^{18y}}\).
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
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
17 Jul 2016, 10:49
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abhilash53
\(28^{18y}\);
\((4*7)^{18y}\);
\((2^2*7)^{18y}\);
\((2^2)^{18y}*7^{18y}\);
\(2^{36y}*7^{18y}\).
Hope it's clear.
