Carcass wrote:
If \(3^x+3^x+3^x=3^{21}\), what is x?
(A) 1
(B) 3
(C) 7
(D) 20
(E) 21
ASIDE-----------------------------
Notice the following:
k + k + k =
3k
w² + w² + w² =
3w²
xy + xy + xy =
3xy
Likewise, 3^x + 3^x + 3^x =
3(3^x)
-------------------------------------
Now let's answer the question....
Given: \(3^x+3^x+3^x=3^{21}\)
Simplify left side: \(3(3^x)=3^{21}\)
Rewrite the first \(3\) as follows: \((3^1)(3^x)=3^{21}\)
Apply the Product Law to simplify the left side: \(3^{x+1}=3^{21}\)
Now that we have the same base on both sides, we can conclude that the exponents are equal: \(x+1 = 21\)
Solve to get: \(x = 20\)
Answer: D
Cheers,
Brent