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Solve for x :
256^(x - 2) = 2^(-x - 7)

-256^(x - 2) = (-1)^1·2^(8 x - 16):
(-1)^1·2^(-x - 7) = (-1)^1·2^(8 x - 16)

Equate exponents of -1 and 2 on both sides:
1 = 1 and -x - 7 = 8 x - 16

All equations give x = 1 as the solution:
 
x = 1

I'm not sure where you learned this, but there is a lot of extraneous math in your solution. All the "1^ -1" is completely unnecessary and confusing, although it does somehow give the correct solution. Here's a simpler way to do it:

256^x-2 = 2^-x-7 | 256 is equal to 2⁸, so you can make the bases the same

2^8x-16=2^-x-7 | Now, you can take away the bases and solve algebraically. If you perform log₂ on both sides, you're left with just the exponents, so I'm just not showing a step.

8x-16=-x-7 | Next, you solve algebraically.

8x+x-16+16=-x+x-7+16

9x=9 | Divide both sides by nine.

x=1 | You have your answer!

 

Now, let's plug it back in to check if it works.

 

256^x-2 = 2^-x-7 | Replace x with our answer, 1

256^1-2=2^-1-7 | Change 256 to 2⁸

2^8-16=2^-1-7 | Simplify

2^-8=2^-8

 

It works! It may look hard at first, but once you try it out it's much easier and saves you the extra steps.

 

TLDR; Match your bases, take the log and solve algebraically.

2^8 is 256 , so this is equivalent to:

2^8(x-2) =2^{(-x-7)    Now equate the exponents}

 

8(x-2) = -x-7

8x-16 = -x-7

9x = 9

x=1