# Find the value of x for the inequality to hold. Log 3 (x^2 + 1) =< log 3 (2x + 5)

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log 3 (x^2 + 1) =< log 3 (2x + 5)

since log is an increasing function , then:

if log a < log b , then a < b

==> x^2 + 1 = < 2x + 5

Move all terms to the left side:

==> x^2 - 2x - 4 = < 0

x1 = [2 + sqrt(4+16)/2 = 1+ sqrt5

x2= 1-sqrt5 (impossible)

==> Then x belongs to [1-sqrt5, 1+sqrt5]

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].**

log3 (x^2+1) = log3 (2x+5)

To find x for which the above holds.

Solution:

log3 (x^2+1) =< log3 (2x+5).

Since the bases are same , the inequality holds for antilog. Therefore,

x^2+1 = < 2x+1 ,

x^2 < 2x. Or

x^2-2x = < 0.

x(x-2) = < 0 .

The above product is negative iff 0 < x < 2.

So x should belong to the open interval ( 0 , 2) in order that the given inequality holds good.