The task is to find the value of x such that:

### 2^ (x -3) - 8 = 0

We make use of the laws of indices to separate the "x-3" index into 2 terms:

2^x . 2^(-3) - 8 = 0

Add 8 on both sides:

2^x . 2^(-3) = 8

2^x . 1/ 2^(3) = 8

We know that 2^3 = 8:

2^x . 1/8 = 8

Multiply by 8 on both sides:

2^x = 64

We also know that 64 can be written as 2^6

2^x = 2^6

Comparing the indices,

** x=6**

Counter check

It is a good habit to counter check our answer by putting the values back into the original equation:

### LHS = 2^ (6 -3) - 8 = 2^3 -8 = 0 = RHS

**Therefore, we confirm that x=6 is the answer**

You have provided the equation to be solved as 22^ (x -3) - 8 = 0.

I think it should be 2^ (x -3) - 8 = 0, and have changed the question accordingly.

=> 2^ (x-3) - 8 =0

Now add 8 to both the sides

=> 2^(x-3) = 8

express 8 as 2^3

=> 2^(x-3) = 2^3

we can now equate x-3 and 3

=> x-3 = 3

=> x = 6

**Therefore the solution for 2^ (x -3) - 8 = 0 is x = 6.**

Since we have 2^(x-3), we'll apply the quotient rule:

a^(b-c) = a^b/a^c

We'll put a = 2, b = x and c = 3

2^(x-3) = 2^x/2^3

But 2^3 = 8

2^(x-3) = 2^x/8

We'll re-write the equation:

2^x/8 - 8 = 0

We'll multiply by 8 both sides:

2^x - 64 = 0

We'll add 64 both sides:

2^x = 64

We'll write 64 as a power of 2:

64 = 2^6

2^x = 2^6

Since the bases are matching, we'll apply one to one property:

**x = 6**

To solve 2^(x-3) -8 = 0.

We add 8 to both sides:

2^(x-3) = 8. We write 2^3 for 8 on the right side.

2^(x-3) = 2^3.

Since both sides of the equation has exponents to the same base 2, we equate the exponents of 2 of both sides:

x-3 = 3.

Add 3 to both sides:

x = 3+3 = 6.

Therefore x = 6 is the solution of 2^(x-3) - 8 = 0.