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If x^a = 2, x^b = 3 and x^c = 6 then x^(3a + b - 2c) = ?
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15 Apr 2020, 08:37
15 Apr 2020, 08:37
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81% (01:39) correct
18% (01:38) wrong based on 27 sessions
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If \(x^a = 2\), \(x^b = 3\) and\(x^c = 6\) then \(x^{(3a+b-2c)} =\) ?
(A) 1/9
(B) 1/6
(C) 1/2
(D) 2/3
(E) 3/4
(A) 1/9
(B) 1/6
(C) 1/2
(D) 2/3
(E) 3/4
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If x^a = 2, x^b = 3 and x^c = 6 then x^(3a + b - 2c) = ?
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15 Apr 2020, 09:23
15 Apr 2020, 09:23
Carcass wrote:
If \(x^a = 2\), \(x^b = 3\) and\(x^c = 6\) then \(x^{(3a+b-2c)} =\) ?
(A) 1/9
(B) 1/6
(C) 1/2
(D) 2/3
(E) 3/4
(A) 1/9
(B) 1/6
(C) 1/2
(D) 2/3
(E) 3/4
Take: \(x^a = 2\)
Raise both sides of the equation to the power of 3 to get: \((x^a)^3 = 2^3\)
Simplify: \(x^{3a} = 8\)
Take: \(x^c = 6\)
Raise both sides of the equation to the power of 2 to get: \((x^c)^2 = 6^2\)
Simplify: \(x^{2c} = 36\)
So, \((x^{3a})(x^b) = (8)(3)\)
Simplify: \(x^{3a+b}=24\)
So, \(\frac{x^{3a+b}}{x^{2c}}=\frac{24}{36}\)
Simplify: \(x^{3a+b-2x}=\frac{24}{36}\)
Simplify: \(x^{3a+b-2x}=\frac{2}{3}\)
Answer: D
Cheers,
Brent
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Re: If x^a = 2, x^b = 3 and x^c = 6 then x^(3a + b - 2c) = ?
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30 May 2023, 02:31
30 May 2023, 02:31
Hello from the GRE Prep Club BumpBot!
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Thanks to another GRE Prep Club member, I have just discovered this valuable topic, yet it had no discussion for over a year. I am now bumping it up - doing my job. I think you may find it valuable (esp those replies with Kudos).
Want to see all other topics I dig out? Follow me (click follow button on profile). You will receive a summary of all topics I bump in your profile area as well as via email.
