The exponential equation `32^(3x - 2) = 256^x` has to be solved.

For any positive number B, if `B^a = B^b` , we can say a = b.

`32^(3x - 2) = 256^x`

Notice that `32 = 2^5` and `256 = 2^8` . The equation can be rewritten as:

`(2^5)^(3x - 2) = (2^8)^x`

Use the property `(a^b)^c = a^(b*c)`

`2^(5*(3x - 2)) = 2^(8*x)`

`2^(15x - 10) = 2^(8x)`

As the base 2 is positive, the exponents can be equated.

15x - 10 = 8x

7x = 10

x = `10/7`

The solution of the equation `32^(3x - 2) = 256^x` is `x = 10/7`

This is simple and is as follows

`32^(3x - 2) = 256^(x)`

To solve such questions we need to use logarithims for simple calculations

`so`

`=> Log(32^(3x-2))=log(256^x) `

using the identity `log(a^b) = b*log(a)`

`=> (3x-2)log(32)=x(log(256)) `

`=> (3x-2)log(2^5) = x(log(2^8)`

`=> (3x-2)*5*log(2)=x*8*(log(2)) `

`=> 5(3x-2)=8x `

`=>15x-10=8x `

`=> 15x-8x=10 `

`=> 7x=10 `

`=> x=10/7 `

is the solution of x . simple :)