By logarithm definition, yields:

`log_a x = b => x = a^b`

Reasoning by analogy, yields:

`log_3 (2x^2-3x+1) = 0 => 2x^2-3x+1 = 3^0`

Since `3^0 = 1` , yields:

`2x^2-3x+1 = 1 => 2x^2 - 3x + 1 - 1 = 0`

Reducing duplicate members, yields:

`2x^2 - 3x = 0`

Factoring out x yields:

`x(2x - 3) = 0`

Using the zero product rule yields:

`{(x = 0),(2x - 3 = 0):} => {(x = 0),(x = 3/2):}`

You need to test the values above in equation, such that:

`x = 0 => log_3 (2*0^2-3*0+1) = log_3 1 = 0 => 1 = 3^0 => 1 = 1`

`x = 3/2 => log_3 (2*9/4-3*3/2+1) = log_3(9/2-9/2+1) = log_3 1 = 0 => 1 = 1`

**Hence, testing both values in equation, the equation holds, thus, the solutions to the given equation are `x = 0` and **`x = 3/2.`

log base 3(2x^2-3x+1)=0

3^0=2x^2-3x+1

1=2x^2-3x+1

2x^2-3x+1-1=0

2x^2-3x=0

x(2x-3)=0

x=0

OR

2x-3=0

2x=3

x=3/2