If $2^{(x+y)}= 64$ and $128^{(x-y)} = 2$, then what is the value of $x$?
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → 43/14
Given:
a. $2^{(x+y)}= 64$ or $2^{(x+y)}= 2^6$ , So, x+y = 6
b. $128^{(x-y)} = 2$ or $(2^7)^{(x-y)} = 2^1$
So, $7(x - y) = 1$ or $x - y = \frac{1}{7}$
Solve the two resulting linear equations:
$x + y = 6$
$x - y = \frac{1}{7}$
$2x = 6 + \frac{1}{7}$
$x = \frac{43}{14}$