Carcass wrote:
If \(\frac{1}{2^m} + \frac{1}{2^m} = \frac{1}{2^x}\), then x expressed in terms of m is
A. \(\frac{m}{2}\)
B. \(m - 1\)
C. \(m + 1\)
D. \(2m\)
E. \(m^2\)
Kudos to the right answer and explanation
Some examples of an important concept:
x + x =
2x
w² + w² =
2w²
y³z + y³z =
2y³z
10.3 + 10.3 =
2(10.3)
The same applies to \(\frac{1}{2^m} + \frac{1}{2^m}\)
GIVEN: \(\frac{1}{2^m} + \frac{1}{2^m} = \frac{1}{2^x}\)
Apply property to get: \(2(\frac{1}{2^m})=\frac{1}{2^x}\)
Rewrite left side as: \(\frac{2}{2^m}=\frac{1}{2^x}\)
This is the same as: \(\frac{2^1}{2^m}=\frac{2^0}{2^x}\)
Apply Quotient Law to both sides to get: \(2^{1-m} = 2^{0-x}\)
Since both sides have the same base, we can conclude that their exponents are equal: \(1-m = 0-x\)
Simplify: \(1-m = -x\)
Multiply both side by -1 to get: \(-1+m = x\)
Rewrite as follows: \(m-1 = x\)
Answer: B
Cheers,
Brent
_________________
Brent Hanneson - founder of Greenlight Test Prep