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
2x + 5 > 0
We'll add -5 both sides:
We'll divide by 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
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
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.
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.