If $2^{(x+y)}= 64$ and $128^{(x-y)} = 2$, then what is the value of $x$?

43/14

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:

$x + y = 6$

$x - y = \frac{1}{7}$

$2x = 6 + \frac{1}{7}$

$x = \frac{43}{14}$​