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
\(6^{-2}=\frac{1}{6^2}=\frac{1}{36}\)
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