Find the value of x for the inequality to hold. Log 3 (x^2 + 1) =

We'll start by imposing the constraints of existance of
logarithm function.

x^2 + 1>0, which is true for any
value of x

and

2x + 5 >
0

We'll add -5 both
sides:

2x>-5

We'll
divide by
2:

x>-5/2

So, for the
logarithms to exist, the values of x have to belong to the interval (-5/2,
+inf.)

Now, we'll solve the inequality. For the beginning,
we notice that the bases of logarithms are matching and they are >1, so the
direction of the inequality remains unchanged, if we'll apply the one to one property of
logarithms:

x^2 + 1 =<2x +
5

We'll move all terms to one
side:

x^2 - 2x + 1 - 5 =<
0

x^2 - 2x - 4 =< 0

To
solve the inequality above, first we have to calculate the roots of the equation x^2 -
2x - 4 = 0.

After that, we'll write the expression in a
factored form as:

1*(x-x1)(x-x2) =<
0

So, let's apply the quadratic formula to calculate the
roots:

x1 =
[2+sqrt(4+16)]/2

x1 =
(2+2sqrt5)/2

x1 =
2(1+sqrt5)/2

x1 =
1+sqrt5

x2 =
1-sqrt5

The inequality will be written
as:

(x - 1 - sqrt5)(x - 1 + sqrt5 ) =<
0

Now, we'll discuss the
inequality:

- the product is  negative if one factor is
positive and the other is negative:

x - 1 - sqrt5 >=
0

We'll add 1 + sqrt5 both
sides:

x > = 1 +
sqrt5

and

x - 1 + sqrt5
=< 0

x =< 1 -
sqrt5

The common solution is the empty
set.

Now, we'll consider the other
alternative:

x - 1 - sqrt5  =<
0

x =< 1 +
sqrt5

and

x - 1 +
sqrt5 >= 0

x >= 1 -
sqrt5

So, x belongs to the interval [1 - sqrt5 , 1 +
sqrt5].

Finally, the solution of the
inequality is the inetrval identified above: [1 - sqrt5 , 1
+ sqrt5].