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07 Sep 2017, 23:23
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If \(3^x = 6^{2x}*\frac{1}{m}\), which of the following is equivalent to m?
A. \(\frac{1}{4}\)
B. \(2^x\)
C. \(6^x\)
D. \(3^x2^{2x}\)
E. \(14\)
A. \(\frac{1}{4}\)
B. \(2^x\)
C. \(6^x\)
D. \(3^x2^{2x}\)
E. \(14\)
Updated on: 08 Sep 2017, 01:09
Originally posted by sarathgopinath on 08 Sep 2017, 01:02.
Last edited by sarathgopinath on 08 Sep 2017, 01:09, edited 1 time in total.
Last edited by sarathgopinath on 08 Sep 2017, 01:09, edited 1 time in total.
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Bunuel
If \(3^x = 6^{2x}*\frac{1}{m}\), which of the following is equivalent to m?
A. \(\frac{1}{4}\)
B. \(2^x\)
C. \(6^x\)
D. \(3^x2^{2x}\)
E. \(14\)
A. \(\frac{1}{4}\)
B. \(2^x\)
C. \(6^x\)
D. \(3^x2^{2x}\)
E. \(14\)
\(3^x = 6^{2x}*\frac{1}{m}\)
\({3^{x}} * m = (3*2)^{2x}\)
\({3^{x}} *m = 3^{2x}\)*\(2^{2x}\)
Now comparing both the sides, m should be equal to \(3^{x}*2^{2x}\)
(D)
08 Sep 2017, 01:08
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Given: \(3^x = 6^{2x}*\frac{1}{m}\)
Since, we have been asked to find the value of m, we can re-write the equation as \(m = \frac{6^{2x}}{3^x}\)
Rules:
\(a^m = b^m*c^m\) (when \(a=b*c\))
\(\frac{a^{m}}{a^{n}} = a^{m-n}\)
Therefore, m can be simplified further as follows:
\(m = \frac{3^{2x}*2^{2x}}{3^x} = (2^{2x})*\frac{3^{2x}}{3^x} = 2^{2x}*3^{2x-x} = 3^x*2^{2x}\)(Option D)
Since, we have been asked to find the value of m, we can re-write the equation as \(m = \frac{6^{2x}}{3^x}\)
Rules:
\(a^m = b^m*c^m\) (when \(a=b*c\))
\(\frac{a^{m}}{a^{n}} = a^{m-n}\)
Therefore, m can be simplified further as follows:
\(m = \frac{3^{2x}*2^{2x}}{3^x} = (2^{2x})*\frac{3^{2x}}{3^x} = 2^{2x}*3^{2x-x} = 3^x*2^{2x}\)(Option D)
