You need to solve for x the given logarithmic equation, such that:

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

Reducing duplicate terms yields:

`2 = -3x + 5`

You need to isolate to one side the terms that contain x, such that:

`3x = -2 + 5 => 3x = 3 => x = 3/3 => x = 1`

You need to test back the value x = 1 in equation, such that:

`log(1^2+2) = log (1^2-3+5) => log3 = log3`

**Hence, evaluating the solution to the given logarithmic equation, yields **`x = 1.`

The equation log(x^2+2)=log (x^2-3x+5) has to be solved for x.

From the given equation a quadratic equation can be obtained by equating x^2 + 2 and x^2 - 3x + 5

x^2 + 2 = x^2 - 3x + 5

-3x = -3

x = 1

The solution of the equation log(x^2+2)=log (x^2-3x+5) is x = 1

We'll impose the constraints of existence of logarithms.

The first condition is:

x^2 + 2 > 0

Since x^2 is always positive for any value of x, the amount:

x^2 + 2 is also positive, fro any value of x.

The second condition is:

x^2 - 3x + 5 > 0

We'll calculate the discriminant of the quadratic to verify if it is negative:

delta = 9 - 20 = -11 < 0

Since delta is negative, the expression x^2 - 3x + 5 is also positive, fro any value of x.

Conclusion: The solution of the equtain could be any value of x.

Now, we'll solve the equation:

log( x^2 + 2 ) = log ( x^2 - 3x + 5)

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

x^2 + 2 = x^2 - 3x + 5

We'll eliminate and combine like terms:

3x - 3 = 0

We'll divide by 3:

x - 1 = 0

x = 1

The solution of the equation is x = 1.