*Solve the following inequations:-*

1.log x^2 to the base(2*x+3)<1

2.|x^2+3*x|+x^2-2>=0

3.(2*x+5)^1/2+(x-1)^1/2>8

log( x^2 )[(2*x+3)] < 1

Let log( x^2 )[(2*x+3)] = p ----- (1)

i.e. p < 1

Then, by definition of logs, using (1),

(2x + 3)^p = x^2

And,

(2x + 3)^p < (2x + 3)^1 (using p < 1, assuming (2x+3) > 1)

x^2 < (2x + 3)^1

x^2 – 2x – 3 < 0

(x – 3)(x + 1) < 0

x < 3 and x > -1

**Answer: [-1 < x < 0], [0 < x < 3]**

|x^2+3*x| + x^2 - 2 >= 0

|x^2+3*x| >= -x^2 + 2

i.e.

(x^2+3*x) >= -x^2 + 2 -(x^2+3*x) >= -x^2 + 2

2x^2 + 3x – 2 >= 0 -3x – 2 >= 0

(2x – 1)(x + 2) >= 0 3x <= -2

x >= ½ , x <= -2 x <= -2/3

x <= -2 and x <= -2/3 is satisfied by x <= -2/3

**Answer: [-∞, -2/3], [1/2, ∞]**

(2*x + 5)^1/2 + (x - 1)^1/2 > 8

Squaring both sides,

(2*x + 5) + (x - 1) + 2*{(2*x + 5)(x - 1)}^1/2 > 64

3x + 4 + 2*{(2*x + 5)(x - 1)}^1/2 > 64

2*{(2*x + 5)(x - 1)}^1/2 > 64 – 3x – 4

2*{(2*x + 5)(x - 1)}^1/2 > 60 – 3x

Squaring both sides,

4{2x^2 + 3x - 5} > 3600 – 360x + 9x^2

8x^2 + 12x - 20 > 3600 – 360x + 9x^2

x^2 – 372x + 3620 < 0

(x – 10)(x – 362) < 0

[10 < x < 362]

The above range is a solution to the final inequality, but the initial inequality is satisfied with [x > 10] since square root function only considers positive values.

**Answer: [x > 10]**