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

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

We know that if log a = log b

Then a = b

Then we know that:

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

Reduce similar terms (x^2):

==> 2 = -3x + 5

Now solve the equation:

Move -3x to the left side and 2 to the right side:

==> 3x = 5-2

==> 3x = 3

Now , divide by 3:

==> x= 1

Now let us check the answer.

Subsitute x=1 in the equality.

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

log (1+ 2) = log (1-3+5)

log 3 = log 3

**Then the answer 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.**

To solve log(x^2+2) = log(x^2-3x+5).

Since both sides are logarithms, we take anti logarithms of both side:

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

x^2+2 - (x^2-3x+5) = 0.

(x^2-x^2)+2+3x-5 = 0.

3x+2-5 = 0.

3x-3 = 0.

3(x-1) = 0.

We divide both sides by 3.

x -1 = 0.

So x= 1.