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:

1. $2(x + y) = 64$
2. $128(x - y) = 2$

Step 1: Simplify both equations

From (1):

$x + y = 32 \quad \text{(Equation A)}$

From (2):

$x - y = \frac{2}{128} = \frac{1}{64} \quad \text{(Equation B)}$

Step 2: Add Equation A and B

$(x+y) + (x-y) = 32 + \frac{1}{64}$

$2x = \frac{2048 + 1}{64} = \frac{2049}{64}$

$x = \frac{2049}{128}$

Step 3: Convert to given options

$\frac{2049}{128} = \frac{43}{14}$