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28 Jul 2020, 23:39
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D
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
65%
(hard)
Question Stats:
65% (02:54) correct
35%
(02:52)
wrong
based on 327
sessions
History
Date
Time
Result
Not Attempted Yet
If \((2^{2x})(3^{x−1})(1/4^{x−2})(27^{1/x})=(6^{x+2})(1/2)(9^{x−4})\), what is the value of x ?
A. -3
B. 1
C. 2
D. 3
E. 6
A. -3
B. 1
C. 2
D. 3
E. 6
29 Jul 2020, 05:34
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buan15
If \((2^{2x})(3^{x−1})(1/4^{x−2})(27^{1/x})=(6^{x+2})(1/2)(9^{x−4})\), what is the value of x ?
A. -3
B. 1
C. 2
D. 3
E. 6
A. -3
B. 1
C. 2
D. 3
E. 6
The best way forward is to solve for individual components to be broken up and then put them back into the final equation.
The following rules of exponents are used:
(1) \(a^m\) * \(a^n\) = \(a^{m + n}\)
(2) \(\frac{a^m }{ a^n}\) = \(a^{m - n}\)
(3) \((ab)^m\) = \(a^m\) * \(b^m\)
(4) \((a^m)^n\) = \(a^{mn}\)
\((2^{2x})(3^{x−1})(1/4^{x−2})(27^{1/x})=(6^{x+2})(1/2)(9^{x−4})\)
\(3^{x - 1}\) = \(\frac{3^x}{3}\)
\(\frac{1}{4^{x−2}}\) = \(\frac{1}{\frac{4^x}{4^2}}\) = \(\frac{4^2}{(2^2)^x}\) = \(\frac{16}{2^{2x}}\)
\(27^{1/x}\) = \((3^3)^{\frac{1}{x}}\) = \(3^{\frac{3}{x}}\)
\(6^{x + 2}\) = \(6^x\) * \(6^2\) = \((3 * 2)^x\) * 36 = \(3^x \)* \(2^x\) * 36
\(9^{x - 4}\) = \((3^2)^{x -4}\) = \(3^{2x - 8}\) = \(\frac{3^{2x}}{3^8}\)
Putting these terms into the original expression, we get
\(2^{2x}\) * \(\frac{3^x}{3}\) * \(\frac{16}{2^{2x}}\) * \(3^{\frac{3}{x}}\) = \(3^x \)* \(2^x\) * 36 * \(\frac{1}{2}\) * \(\frac{3^{2x}}{3^8}\)
\(2^{2x}\) and \(3^x\) are cancelled out. Bringing all the constant values to one side, we get
\(\frac{2^x * 3^x * 3^{2x} }{ 3^{\frac{3}{x}}* 3^x}\) = \(\frac{16 * 2 * 3^8}{ 36 * 3} \)
\(2^x\) * \(3^{2x - {\frac{3}{x}}\) = \(\frac{16 * 3^8}{ 18 * 3} \)
\(2^x\) * \(3^{2x - {\frac{3}{x}}\) = \(\frac{2^4 * 3^7}{ 2 * 9} \)
\(2^x\) * \(3^{2x - {\frac{3}{x}}\) = \(\frac{2^3 * 3^8}{ 3^2 * 3} \)
\(2^x\) * \(3^{2x - {\frac{3}{x}}\) = \(2^3 * 3^5 \)
Since we have \(2^x\) = \(2^3\) , therefore x = 3
Option D
Arun Kumar
29 Jul 2020, 07:18
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CrackVerbalGMAT
buan15
If \((2^{2x})(3^{x−1})(1/4^{x−2})(27^{1/x})=(6^{x+2})(1/2)(9^{x−4})\), what is the value of x ?
A. -3
B. 1
C. 2
D. 3
E. 6
A. -3
B. 1
C. 2
D. 3
E. 6
The best way forward is to solve for individual components to be broken up and then put them back into the final equation.
The following rules of exponents are used:
(1) \(a^m\) * \(a^n\) = \(a^{m + n}\)
(2) \(\frac{a^m }{ a^n}\) = \(a^{m - n}\)
(3) \((ab)^m\) = \(a^m\) * \(b^m\)
(4) \((a^m)^n\) = \(a^{mn}\)
\((2^{2x})(3^{x−1})(1/4^{x−2})(27^{1/x})=(6^{x+2})(1/2)(9^{x−4})\)
\(3^{x - 1}\) = \(\frac{3^x}{3}\)
\(\frac{1}{4^{x−2}}\) = \(\frac{1}{\frac{4^x}{4^2}}\) = \(\frac{4^2}{(2^2)^x}\) = \(\frac{16}{2^{2x}}\)
\(27^{1/x}\) = \((3^3)^{\frac{1}{x}}\) = \(3^{\frac{3}{x}}\)
\(6^{x + 2}\) = \(6^x\) * \(6^2\) = \((3 * 2)^x\) * 36 = \(3^x \)* \(2^x\) * 36
\(9^{x - 4}\) = \((3^2)^{x -4}\) = \(3^{2x - 8}\) = \(\frac{3^{2x}}{3^8}\)
Putting these terms into the original expression, we get
\(2^{2x}\) * \(\frac{3^x}{3}\) * \(\frac{16}{2^{2x}}\) * \(3^{\frac{3}{x}}\) = \(3^x \)* \(2^x\) * 36 * \(\frac{1}{2}\) * \(\frac{3^{2x}}{3^8}\)
\(2^{2x}\) and \(3^x\) are cancelled out. Bringing all the constant values to one side, we get
\(\frac{2^x * 3^x * 3^{2x} }{ 3^{\frac{3}{x}}* 3^x}\) = \(\frac{16 * 2 * 3^8}{ 36 * 3} \)
\(2^x\) * \(3^{2x - {\frac{3}{x}}\) = \(\frac{16 * 3^8}{ 18 * 3} \)
\(2^x\) * \(3^{2x - {\frac{3}{x}}\) = \(\frac{2^4 * 3^7}{ 2 * 9} \)
\(2^x\) * \(3^{2x - {\frac{3}{x}}\) = \(\frac{2^3 * 3^8}{ 3^2 * 3} \)
\(2^x\) * \(3^{2x - {\frac{3}{x}}\) = \(2^3 * 3^5 \)
Since we have \(2^x\) = \(2^3\) , therefore x = 3
Option D
Arun Kumar
I got it right but took more than 3 minutes - wondering whether timing is ok or not.
Additionally if we solve for power of 3 then we get 2 values of x: x=3, -1/2.
Am curious if we need to back solve & recheck in this type of scenarios.
Please suggest. Thanks!
