Carcass wrote:
If \(2^k + 2^{k-2} - 2^{k-3} = 2^{2k}*3^m\), what is the value of k+m?
A. -1
B. 0
C. 1
D. 2
E. 3
Given: \(2^k + 2^{k-2} - 2^{k-3} = 2^{2k}*3^m\)
Factor to get: \(2^{k-3}(2^3 + 2^1 - 1) = 2^{2k}*3^m\)
Evaluate: \(2^{k-3}(9) = 2^{2k}*3^m\)
Rewrite \(9\) as a power of \(3\) to get: \(2^{k-3}(3^2) = 2^{2k}*3^m\)
We can now conclude two things: \(2^{k-3}= 2^{2k}\) and \(3^2 = 3^m\)
If \(2^{k-3}= 2^{2k}\), then \(k-3= 2k\), which means \(k = -3\)
If \(3^2 = 3^m\), then \(m = 2\)
So, \(k+m = (-3) + 2 = -1\)
Answer: A
Cheers,
Brent