Mister Exam
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How to use it?
- Inequation:
- log10(x^2-2x-2)<=0
- 1\6x<-12
- log2((x-1)(x^2+3))<=log2(4*x-x^2-3)+log2(5-x)
-
log1/7(x)>=x-8
- Canonical form:
- =0
-
Identical expressions
- log1 zero (x^ two - two x-2)<=0
- logarithm of 10(x squared minus 2x minus 2) less than or equal to 0
- logarithm of 1 zero (x to the power of two minus two x minus 2) less than or equal to 0
- log10(x2-2x-2)<=0
- log10x2-2x-2<=0
- log10(x²-2x-2)<=0
- log10(x to the power of 2-2x-2)<=0
- log10x^2-2x-2<=0
- log10(x^2-2x-2)<=O
-
Similar expressions
- log10(x^2+2x-2)<=0
- log10(x^2-2x+2)<=0
log10(x^2-2x-2)<=0 inequation
The solution
You have entered
[src]
/ 2 \
log\x - 2*x - 2/
----------------- <= 0
log(10)
$$\frac{\log{\left(\left(x^{2} - 2 x\right) - 2 \right)}}{\log{\left(10 \right)}} \leq 0$$
log(x^2 - 2*x - 2)/log(10) <= 0
Detail solution
Given the inequality:
$$\frac{\log{\left(\left(x^{2} - 2 x\right) - 2 \right)}}{\log{\left(10 \right)}} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(\left(x^{2} - 2 x\right) - 2 \right)}}{\log{\left(10 \right)}} = 0$$
Solve:
$$x_{1} = -1$$
$$x_{2} = 3$$
$$x_{1} = -1$$
$$x_{2} = 3$$
This roots
$$x_{1} = -1$$
$$x_{2} = 3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$\frac{\log{\left(\left(x^{2} - 2 x\right) - 2 \right)}}{\log{\left(10 \right)}} \leq 0$$
$$\frac{\log{\left(-2 + \left(\left(- \frac{11}{10}\right)^{2} - \frac{\left(-11\right) 2}{10}\right) \right)}}{\log{\left(10 \right)}} \leq 0$$
but
Then
$$x \leq -1$$
no execute
one of the solutions of our inequality is:
$$x \geq -1 \wedge x \leq 3$$
$$\frac{\log{\left(\left(x^{2} - 2 x\right) - 2 \right)}}{\log{\left(10 \right)}} \leq 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(\left(x^{2} - 2 x\right) - 2 \right)}}{\log{\left(10 \right)}} = 0$$
Solve:
$$x_{1} = -1$$
$$x_{2} = 3$$
$$x_{1} = -1$$
$$x_{2} = 3$$
This roots
$$x_{1} = -1$$
$$x_{2} = 3$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$-1 + - \frac{1}{10}$$
=
$$- \frac{11}{10}$$
substitute to the expression
$$\frac{\log{\left(\left(x^{2} - 2 x\right) - 2 \right)}}{\log{\left(10 \right)}} \leq 0$$
$$\frac{\log{\left(-2 + \left(\left(- \frac{11}{10}\right)^{2} - \frac{\left(-11\right) 2}{10}\right) \right)}}{\log{\left(10 \right)}} \leq 0$$
/141\ log|---| \100/ <= 0 -------- log(10)
but
/141\ log|---| \100/ >= 0 -------- log(10)
Then
$$x \leq -1$$
no execute
one of the solutions of our inequality is:
$$x \geq -1 \wedge x \leq 3$$
_____
/ \
-------•-------•-------
x1 x2
Solving inequality on a graph
Rapid solution
[src]
/ / ___\ / ___ \\ Or\And\-1 <= x, x < 1 - \/ 3 /, And\x <= 3, 1 + \/ 3 < x//
$$\left(-1 \leq x \wedge x < 1 - \sqrt{3}\right) \vee \left(x \leq 3 \wedge 1 + \sqrt{3} < x\right)$$
((x <= 3)∧(1 + sqrt(3) < x))∨((-1 <= x)∧(x < 1 - sqrt(3)))
Rapid solution 2
[src]
___ ___ [-1, 1 - \/ 3 ) U (1 + \/ 3 , 3]
$$x\ in\ \left[-1, 1 - \sqrt{3}\right) \cup \left(1 + \sqrt{3}, 3\right]$$
x in Union(Interval.Ropen(-1, 1 - sqrt(3)), Interval.Lopen(1 + sqrt(3), 3))