Mister Exam
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How to use it?
- Inequation:
- log_(7)((2x-6):(2x-1))>0
- x^log625(2+x)>=log5(x^2+4x+4)
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((2-x^2)*(x-3)^3)/((x+1)*(x^2-3x-4))>=0
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6x^2+17<0
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Identical expressions
- log_(seven)((2x- six):(2x- one))> zero
- logarithm of _(7)((2x minus 6):(2x minus 1)) greater than 0
- logarithm of _(seven)((2x minus six):(2x minus one)) greater than zero
- log_72x-6:2x-1>0
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Similar expressions
- log_(7)((2x-6):(2x+1))>0
- log_(7)((2x+6):(2x-1))>0
log_(7)((2x-6):(2x-1))>0 inequation
The solution
You have entered
[src]
/2*x - 6\ log|-------| \2*x - 1/ ------------ > 0 log(7)
$$\frac{\log{\left(\frac{2 x - 6}{2 x - 1} \right)}}{\log{\left(7 \right)}} > 0$$
log((2*x - 1*6)/(2*x - 1*1))/log(7) > 0
Detail solution
Given the inequality:
$$\frac{\log{\left(\frac{2 x - 6}{2 x - 1} \right)}}{\log{\left(7 \right)}} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(\frac{2 x - 6}{2 x - 1} \right)}}{\log{\left(7 \right)}} = 0$$
Solve:
This equation has no roots,
this inequality is executed for any x value or has no solutions
check it
subtitute random point x, for example
$$x_0 = 0$$
$$\frac{\log{\left(\frac{\left(-1\right) 6 + 2 \cdot 0}{\left(-1\right) 1 + 2 \cdot 0} \right)}}{\log{\left(7 \right)}} > 0$$
so the inequality is always executed
$$\frac{\log{\left(\frac{2 x - 6}{2 x - 1} \right)}}{\log{\left(7 \right)}} > 0$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(\frac{2 x - 6}{2 x - 1} \right)}}{\log{\left(7 \right)}} = 0$$
Solve:
This equation has no roots,
this inequality is executed for any x value or has no solutions
check it
subtitute random point x, for example
$$x_0 = 0$$
$$\frac{\log{\left(\frac{\left(-1\right) 6 + 2 \cdot 0}{\left(-1\right) 1 + 2 \cdot 0} \right)}}{\log{\left(7 \right)}} > 0$$
log(6) ------ > 0 log(7)
so the inequality is always executed
Solving inequality on a graph
Rapid solution
[src]
And(-oo < x, x < 1/2)
$$-\infty < x \wedge x < \frac{1}{2}$$
(-oo < x)∧(x < 1/2)
Rapid solution 2
[src]
(-oo, 1/2)
$$x\ in\ \left(-\infty, \frac{1}{2}\right)$$
x in Interval.open(-oo, 1/2)