We have to solve 3^(x-2)-9=0

3^(x-2)-9=0

=> 3^(x - 2) = 9

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

as the base is same, equate the exponent.

x - 2 = 2

=> x = 4

**The solution of the equation is x = 4**

The equation 3^(x-2)-9=0 has to be solved.

If b^x = b^y, we can equate x and y and if a^x = b^x we can equate a and b.

Here 3^(x-2)-9=0, this can be written as:

3^(x-2) = 9

Now 9 = 3^2

3^(x - 2) = 3^2

As the base 3 is common on both the sides, the exponents can be equated

x - 2 = 2

x = 2 + 2 = 4

The solution of 3^(x-2)-9=0 is x = 4

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

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

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

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

But 3^2 = 9

3^(x-2) = 3^x/9

We'll re-write the equation:

3^x/9 - 9 = 0

We'll multiply by 9 both sides:

3^x - 81 = 0

We'll add 81 both sides:

3^x = 81

We'll write 81 as a power of 3:

81 = 3^4

3^x = 3^4

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

x = 4