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If 14/6^-2 * 45 = 2^x*3^4*35
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21 May 2019, 11:21
21 May 2019, 11:21
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Question Stats:
95% (01:27) correct
4% (03:36) wrong based on 41 sessions
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If \(\frac{14}{6^{-2}} * 45 = 2^x*3^4*35\) , what is the value of \(x\) ?
A. -2
B. 1
C. 2
D. 3
E. 4
A. -2
B. 1
C. 2
D. 3
E. 4
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Posts: 186
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Location: United States
GMAT 1: 770 Q51 V44
GRE 1: Q169 V168
WE:Education (Education)
Re: If 14/6^-2 * 45 = 2^x*3^4*35
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21 May 2019, 18:16
21 May 2019, 18:16
1
Expert Reply
Quote:
If \(\frac{14}{6^{-2}} * 45 = 2^x*3^4*35\) , what is the value of \(x\) ?
With textbook arithmetic work step by step and eliminate fractions first. In this case, if there is a negative exponent in a denominator, you can simply move the base to the numerator and make the negative exponent positive.
Therefore, 6^2 * 14 * 45 = 2^x * 3^4 * 35.
Now, break the left side compound product into prime factors to determine the number of factors of 2.
So, 2^2 * 2 * 7 means that the left hand side of the equation includes 2^3 which means on the right hand side of the equation x = 3. Select Choice D.
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Re: If 14/6^-2 * 45 = 2^x*3^4*35
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23 May 2019, 07:37
23 May 2019, 07:37
2
Carcass wrote:
If \(\frac{14}{6^{-2}} * 45 = 2^x*3^4*35\) , what is the value of \(x\) ?
A. -2
B. 1
C. 2
D. 3
E. 4
A. -2
B. 1
C. 2
D. 3
E. 4
\(6^{-2}=\frac{1}{6^2}=\frac{1}{36}\)
So, \(\frac{14}{6^{-2}}=\frac{(\frac{14}{1})}{(\frac{1}{36})}=(\frac{14}{1})(\frac{36}{1})=(14)(36)\)
So, \(\frac{14}{6^{-2}} * 45 = (14)(36)(45)\)
Our equation becomes: \((14)(36)(45)=(2^x)(3^4)(35)\)
Find the prime factorization of remaining values: \((2)(7)(2)(2)(3)(3)(3)(3)(5)=(2^x)(3^4)(5)(7)\)
Rearrange: \((2)(2)(2)(3)(3)(3)(3)(5)(7)=(2^x)(3^4)(5)(7)\)
Rewrite as powers: \((2^3)(3^4)(5)(7)=(2^x)(3^4)(5)(7)\)
We can see that x = 3
Answer: D
Cheers,
Brent
