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  • Inequation:
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  • x+4>5*x
  • (1/3)^(x/2)>9
  • log(3.3)*3*(1-(7/5)*x)<1
  • Canonical form:
  • =0

  • Identical expressions

  • nine ^(x^ two - one)- thirty-six * three ^(x^ two - three)+ three <= zero
  • 9 to the power of (x squared minus 1) minus 36 multiply by 3 to the power of (x squared minus 3) plus 3 less than or equal to 0
  • nine to the power of (x to the power of two minus one) minus thirty minus six multiply by three to the power of (x to the power of two minus three) plus three less than or equal to zero
  • 9(x2-1)-36*3(x2-3)+3<=0
  • 9x2-1-36*3x2-3+3<=0
  • 9^(x²-1)-36*3^(x²-3)+3<=0
  • 9 to the power of (x to the power of 2-1)-36*3 to the power of (x to the power of 2-3)+3<=0
  • 9^(x^2-1)-363^(x^2-3)+3<=0
  • 9(x2-1)-363(x2-3)+3<=0
  • 9x2-1-363x2-3+3<=0
  • 9^x^2-1-363^x^2-3+3<=0
  • 9^(x^2-1)-36*3^(x^2-3)+3<=O

  • Similar expressions

  • 9^(x^2-1)+36*3^(x^2-3)+3<=0
  • 9^(x^2-1)-36*3^(x^2-3)-3<=0
  • 9^(x^2-1)-36*3^(x^2+3)+3<=0
  • 9^(x^2+1)-36*3^(x^2-3)+3<=0
Solving inequalities/ 9^(x^2-1)-36*3^(x^2-3)+3<=0

9^(x^2-1)-36*3^(x^2-3)+3<=0 inequation

A inequation with variable

The solution

You have entered [src] 
  2            2             
 x  - 1       x  - 3         
9       - 36*3       + 3 <= 0
$$\left(- 36 \cdot 3^{x^{2} - 3} + 9^{x^{2} - 1}\right) + 3 \leq 0$$ 
-36*3^(x^2 - 3) + 9^(x^2 - 1) + 3 <= 0
Detail solution
Given the inequality:
$$\left(- 36 \cdot 3^{x^{2} - 3} + 9^{x^{2} - 1}\right) + 3 \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\left(- 36 \cdot 3^{x^{2} - 3} + 9^{x^{2} - 1}\right) + 3 = 0$$
Solve:
$$x_{1} = -1$$
$$x_{2} = 1$$
$$x_{3} = - \sqrt{2}$$
$$x_{4} = \sqrt{2}$$
$$x_{1} = -1$$
$$x_{2} = 1$$
$$x_{3} = - \sqrt{2}$$
$$x_{4} = \sqrt{2}$$
This roots
$$x_{3} = - \sqrt{2}$$
$$x_{1} = -1$$
$$x_{2} = 1$$
$$x_{4} = \sqrt{2}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{3}$$
For example, let's take the point
$$x_{0} = x_{3} - \frac{1}{10}$$
=
$$- \sqrt{2} - \frac{1}{10}$$
=
$$- \sqrt{2} - \frac{1}{10}$$
substitute to the expression
$$\left(- 36 \cdot 3^{x^{2} - 3} + 9^{x^{2} - 1}\right) + 3 \leq 0$$
$$\left(- 36 \cdot 3^{-3 + \left(- \sqrt{2} - \frac{1}{10}\right)^{2}} + 9^{-1 + \left(- \sqrt{2} - \frac{1}{10}\right)^{2}}\right) + 3 \leq 0$$
                        2                          2     
          /  1      ___\             /  1      ___\      
     -1 + |- -- - \/ 2 |        -3 + |- -- - \/ 2 |  <= 0
          \  10        /             \  10        /      
3 + 9                     - 36*3

but
                        2                          2     
          /  1      ___\             /  1      ___\      
     -1 + |- -- - \/ 2 |        -3 + |- -- - \/ 2 |  >= 0
          \  10        /             \  10        /      
3 + 9                     - 36*3

Then
$$x \leq - \sqrt{2}$$
no execute
one of the solutions of our inequality is:
$$x \geq - \sqrt{2} \wedge x \leq -1$$
         _____           _____  
        /     \         /     \  
-------•-------•-------•-------•-------
       x3      x1      x2      x4

Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \geq - \sqrt{2} \wedge x \leq -1$$
$$x \geq 1 \wedge x \leq \sqrt{2}$$
Solving inequality on a graph
Rapid solution [src] 
  /   /               ___\     /   ___              \\
Or\And\1 <= x, x <= \/ 2 /, And\-\/ 2  <= x, x <= -1//
$$\left(1 \leq x \wedge x \leq \sqrt{2}\right) \vee \left(- \sqrt{2} \leq x \wedge x \leq -1\right)$$ 
((1 <= x)∧(x <= sqrt(2)))∨((x <= -1)∧(-sqrt(2) <= x))
Rapid solution 2 [src] 
    ___              ___ 
[-\/ 2 , -1] U [1, \/ 2 ]
$$x\ in\ \left[- \sqrt{2}, -1\right] \cup \left[1, \sqrt{2}\right]$$ 
x in Union(Interval(1, sqrt(2)), Interval(-sqrt(2), -1))
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