log( x^2 + 2 ) = log ( x^2 - 3x + 5) find x value.

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.