# 1) (1/2)^(x^2+x-2)<4^(x-1) 2)(1/2)^(-x^2)+2^((x^2)+3)<18 3)(1/5)^((x^2)-7)-5*0,2^x<0thanx

*print*Print*list*Cite

### 1 Answer

1) You need to use negative property of exponential for left side term such that:

`(1/2)^(x^2+x-2) = (2)^(-x^2-x+2)`

You need to write the base of right side such that:

`4^(x-1) = 2^(2(x-1))`

Since the base 2>1, the exponential function increases as x increases, hence `(2)^(-x^2-x+2) lt 2^(2(x-1)) =gt -x^2-x+2 lt 2(x-1)`

Opening the brackets yields:

`-x^2-x+2 lt 2x - 2`

`-x^2 - 3x + 4 lt 0`

You need to multiply by -1 such that:

`x^2 + 3x - 4 gt 0`

You need to find the roots of equation `x^2 + 3x - 4 = 0` such that:

`x_(1,2) = (-3+-sqrt(9+16))/2 =gt x_(1,2) = (-3+-sqrt25)/2`

`x_(1,2) = (-3+-5)/2 =gt x_1 = 1 ; x_2 = -4`

You should select a value between -4 and 1 and then you need to substitute this value in equation such that:

`0^2 + 3*0 - 4 = -4 lt 0`

**Hence, the inequality holds if `x in (-oo,-4)U(1,oo).` **

2) `(1/2)^(-x^2)+2^((x^2)+3)lt18`

You need to use the exponential properties for `2^((x^2)+3)` such that:

`2^((x^2)+3) = 2^(x^2)*2^3`

`2^(x^2) + 2^(x^2)*2^3 lt 18`

You need to factor out `2^(x^2) ` such that:

`2^(x^2) (1 + 8) lt 18`

`2^(x^2) * 9 lt 18`

You need to divide by 9 both sides such that:

`2^(x^2) lt 2 =gt x^2 lt 1`

**Since the roots of equation `x^2 - 1 = 0` are `x_(1,2) = +-1` , hence the inequality holds for `x in (-1,1).` **

3) `(1/5)^((x^2)-7)-5*0.2^xlt0`

You need to write `0.2^x = (2/10)^x = (1/5)^x`

You should come up with the substitution `(1/5)^x = y ` such that:

`5^7*y^x - 5y lt 0`

You need to divide by 5 such that:

`5^6*y^x - y lt 0`

You need to factor out y such that:

`y(5^6y^(x-1) - 1) lt 0`

Since `y = (1/5)^x lt0 ` expresses a contradiction, then you should solve the inequality `5^6y^(x-1) - 1 lt 0 ` such that:

`5^6y^(x-1) - 1 lt 0 =gt 5^6y^(x-1) lt 1`

You need to divide by `5^6` both sides such that:

`y^(x-1) lt (1/5)^6`

`(1/5)^(x-1) lt (1/5)^6 `

Since the base `0lt1/5lt1` , the exponential function decreases as x increases, hence `(1/5)^(x-1) lt (1/5)^6 =gt x-1gt6 =gt xgt7`

**Hence, the inequality holds for `x in (7,oo).` **