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655-705 (Hard)|   Algebra|   Exponents|      
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buan15
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If (2^2x)(3^x−1)(1/4^x−2)(27^1/x)=(6^x+2)(1/2)(9^x−4), find x? 28 Jul 2020, 23:39
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Difficulty:

55% (hard)

Question Stats:

69% (02:51) correct 31% (02:57) wrong based on 455 sessions

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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
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D
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CrackverbalGMAT
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If (2^2x)(3^x−1)(1/4^x−2)(27^1/x)=(6^x+2)(1/2)(9^x−4), find x? 29 Jul 2020, 05:34
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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
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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

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If (2^2x)(3^x−1)(1/4^x−2)(27^1/x)=(6^x+2)(1/2)(9^x−4), find x? 29 Jul 2020, 07:18
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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

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
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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!
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If (2^2x)(3^x−1)(1/4^x−2)(27^1/x)=(6^x+2)(1/2)(9^x−4), find x? Updated on: 29 Jul 2020, 11:02
Originally posted by CrackverbalGMAT on 29 Jul 2020, 09:40.
Last edited by CrackverbalGMAT on 29 Jul 2020, 11:02, edited 1 time in total.
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buan15


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!
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Hello.

yes, when solving for the power of 3, you get 2 values, but x = -1/2 does not satisfy \(2^x\) = 8. So the same can be eliminated. It is a good practice to put the value of x and check if it satisfies the required conditions.

The question could be done within a minute to a minute and a half. The Maximum time would be to convert the 5 terms into powers of 2 and 3, and move them to the same side. With practice this speed can be achieved.

Trying to plugin the options is another method. We can eliminate options B and E as they would end up as \(\sqrt{27}\) and \(\sqrt{3}\) respectively for \(27^{\frac{1}{3}}\). We would have to still contend with the other 3 options, which could take more time.

Using the concepts would be the best bet in this case.

Hope this helps.
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If (2^2x)(3^x−1)(1/4^x−2)(27^1/x)=(6^x+2)(1/2)(9^x−4), find x? 29 Jul 2020, 10:23
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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

correct:

\((2^{2x})(3^{x−1})(1/4^{x−2})(27^{1/x})=(6^{x+2})(1/2)(9^{x−4}) => (2^{2x})(3^{x−1})(1/2^{2x−4})(3^{3/x})=(2^{x+2})(3^{x+2})(2^{-1})(3^{2x−8}) => (2^{2x-2x+4})(3^{x−1+3/x})=(2^{x+2-1})(3^{x+2+2x-8}) \)

=> \(2^4*3^{x−1+3/x}=2^{x+1}*3^{3x-6} \), now equating power of 2 from both side, we get 4=x+1=> x= 3

E. 6
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If (2^2x)(3^x−1)(1/4^x−2)(27^1/x)=(6^x+2)(1/2)(9^x−4), find x? 07 Oct 2025, 03:55
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