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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.`
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.
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