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How to use it?
- Inequation:
- 3log11(x^2+8x-9)<4+log11(x-1)^3-log11(x+9)
-
tgx>=-1
- logx(4x-3)>=2
- 4(x+5)-40≥-5+x
-
Identical expressions
- three log one 1(x^ two +8x- nine)< four +log11(x-1)^3-log11(x+ nine)
- 3 logarithm of 11(x squared plus 8x minus 9) less than 4 plus logarithm of 11(x minus 1) cubed minus logarithm of 11(x plus 9)
- three logarithm of one 1(x to the power of two plus 8x minus nine) less than four plus logarithm of 11(x minus 1) cubed minus logarithm of 11(x plus nine)
- 3log11(x2+8x-9)<4+log11(x-1)3-log11(x+9)
- 3log11x2+8x-9<4+log11x-13-log11x+9
- 3log11(x²+8x-9)<4+log11(x-1)³-log11(x+9)
- 3log11(x to the power of 2+8x-9)<4+log11(x-1) to the power of 3-log11(x+9)
- 3log11x^2+8x-9<4+log11x-1^3-log11x+9
-
Similar expressions
- 3log11(x^2+8x-9)<4+log11(x-1)^3+log11(x+9)
- 3log11(x^2+8x-9)<4+log11(x-1)^3-log11(x-9)
- 3log11(x^2+8x+9)<4+log11(x-1)^3-log11(x+9)
- 3log11(x^2+8x-9)<4+log11(x+1)^3-log11(x+9)
- 3log11(x^2+8x-9)<4-log11(x-1)^3-log11(x+9)
- 3log11(x^2-8x-9)<4+log11(x-1)^3-log11(x+9)
3log11(x^2+8x-9)<4+log11(x-1)^3-log11(x+9) inequation
The solution
You have entered
[src]
/ 2 \ 3
3*log\x + 8*x - 9/ /log(x - 1)\ log(x + 9)
------------------- < 4 + |----------| - ----------
log(11) \ log(11) / log(11)
$$\frac{3 \log{\left(x^{2} + 8 x - 9 \right)}}{\log{\left(11 \right)}} < \left(\frac{\log{\left(x - 1 \right)}}{\log{\left(11 \right)}}\right)^{3} - \frac{\log{\left(x + 9 \right)}}{\log{\left(11 \right)}} + 4$$
3*log(x^2 + 8*x - 1*9)/log(11) < (log(x - 1*1)/log(11))^3 - log(x + 9)/log(11) + 4
Detail solution
Given the inequality:
$$\frac{3 \log{\left(x^{2} + 8 x - 9 \right)}}{\log{\left(11 \right)}} < \left(\frac{\log{\left(x - 1 \right)}}{\log{\left(11 \right)}}\right)^{3} - \frac{\log{\left(x + 9 \right)}}{\log{\left(11 \right)}} + 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{3 \log{\left(x^{2} + 8 x - 9 \right)}}{\log{\left(11 \right)}} = \left(\frac{\log{\left(x - 1 \right)}}{\log{\left(11 \right)}}\right)^{3} - \frac{\log{\left(x + 9 \right)}}{\log{\left(11 \right)}} + 4$$
Solve:
$$x_{1} = 249.352332301507$$
$$x_{2} = \mathtt{\text{(249.35233230150703+1.2519575680618522e-18j)}}$$
$$x_{3} = \mathtt{\text{(249.35233230150703-2.5108538302468223e-19j)}}$$
$$x_{4} = 2$$
$$x_{5} = \mathtt{\text{(249.35233230150703+1.5537333626017934e-18j)}}$$
Exclude the complex solutions:
$$x_{1} = 249.352332301507$$
$$x_{2} = 2$$
This roots
$$x_{2} = 2$$
$$x_{1} = 249.352332301507$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$1.9$$
substitute to the expression
$$\frac{3 \log{\left(x^{2} + 8 x - 9 \right)}}{\log{\left(11 \right)}} < \left(\frac{\log{\left(x - 1 \right)}}{\log{\left(11 \right)}}\right)^{3} - \frac{\log{\left(x + 9 \right)}}{\log{\left(11 \right)}} + 4$$
$$\frac{3 \log{\left(\left(-1\right) 9 + 1.9^{2} + 8 \cdot 1.9 \right)}}{\log{\left(11 \right)}} < - \frac{\log{\left(1.9 + 9 \right)}}{\log{\left(11 \right)}} + \left(\frac{\log{\left(\left(-1\right) 1 + 1.9 \right)}}{\log{\left(11 \right)}}\right)^{3} + 4$$
one of the solutions of our inequality is:
$$x < 2$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < 2$$
$$x > 249.352332301507$$
$$\frac{3 \log{\left(x^{2} + 8 x - 9 \right)}}{\log{\left(11 \right)}} < \left(\frac{\log{\left(x - 1 \right)}}{\log{\left(11 \right)}}\right)^{3} - \frac{\log{\left(x + 9 \right)}}{\log{\left(11 \right)}} + 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{3 \log{\left(x^{2} + 8 x - 9 \right)}}{\log{\left(11 \right)}} = \left(\frac{\log{\left(x - 1 \right)}}{\log{\left(11 \right)}}\right)^{3} - \frac{\log{\left(x + 9 \right)}}{\log{\left(11 \right)}} + 4$$
Solve:
$$x_{1} = 249.352332301507$$
$$x_{2} = \mathtt{\text{(249.35233230150703+1.2519575680618522e-18j)}}$$
$$x_{3} = \mathtt{\text{(249.35233230150703-2.5108538302468223e-19j)}}$$
$$x_{4} = 2$$
$$x_{5} = \mathtt{\text{(249.35233230150703+1.5537333626017934e-18j)}}$$
Exclude the complex solutions:
$$x_{1} = 249.352332301507$$
$$x_{2} = 2$$
This roots
$$x_{2} = 2$$
$$x_{1} = 249.352332301507$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 2$$
=
$$1.9$$
substitute to the expression
$$\frac{3 \log{\left(x^{2} + 8 x - 9 \right)}}{\log{\left(11 \right)}} < \left(\frac{\log{\left(x - 1 \right)}}{\log{\left(11 \right)}}\right)^{3} - \frac{\log{\left(x + 9 \right)}}{\log{\left(11 \right)}} + 4$$
$$\frac{3 \log{\left(\left(-1\right) 9 + 1.9^{2} + 8 \cdot 1.9 \right)}}{\log{\left(11 \right)}} < - \frac{\log{\left(1.9 + 9 \right)}}{\log{\left(11 \right)}} + \left(\frac{\log{\left(\left(-1\right) 1 + 1.9 \right)}}{\log{\left(11 \right)}}\right)^{3} + 4$$
0.00116959004327434 2.3887627892351
6.85020682073181 4 - ------------------- - ---------------
---------------- < 3 log(11)
log(11) log (11)
one of the solutions of our inequality is:
$$x < 2$$
_____ _____
\ /
-------ο-------ο-------
x_2 x_1Other solutions will get with the changeover to the next point
etc.
The answer:
$$x < 2$$
$$x > 249.352332301507$$
Solving inequality on a graph