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Other calculators
- Square Inequality Step-by-Step
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How to use it?
- Inequation:
- log2(2*x^2-17*x+35)/log7(x+6)-1/log7(x+6)<=4
- log2^2(16+6x-x^2)+10log0,5(16+6x-x^2)+24>0
- tg(5x-p/3)>=-3/3
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tg^2x-4tgx+3>0
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Identical expressions
- log two (two *x^2- seventeen *x+ thirty-five)/log7(x+ six)- one /log7(x+ six)<= four
- logarithm of 2(2 multiply by x squared minus 17 multiply by x plus 35) divide by logarithm of 7(x plus 6) minus 1 divide by logarithm of 7(x plus 6) less than or equal to 4
- logarithm of two (two multiply by x squared minus seventeen multiply by x plus thirty minus five) divide by logarithm of 7(x plus six) minus one divide by logarithm of 7(x plus six) less than or equal to four
- log2(2*x2-17*x+35)/log7(x+6)-1/log7(x+6)<=4
- log22*x2-17*x+35/log7x+6-1/log7x+6<=4
- log2(2*x²-17*x+35)/log7(x+6)-1/log7(x+6)<=4
- log2(2*x to the power of 2-17*x+35)/log7(x+6)-1/log7(x+6)<=4
- log2(2x^2-17x+35)/log7(x+6)-1/log7(x+6)<=4
- log2(2x2-17x+35)/log7(x+6)-1/log7(x+6)<=4
- log22x2-17x+35/log7x+6-1/log7x+6<=4
- log22x^2-17x+35/log7x+6-1/log7x+6<=4
- log2(2*x^2-17*x+35) divide by log7(x+6)-1 divide by log7(x+6)<=4
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Similar expressions
- log2(2*x^2+17*x+35)/log7(x+6)-1/log7(x+6)<=4
- log2(2*x^2-17*x-35)/log7(x+6)-1/log7(x+6)<=4
- log2(2*x^2-17*x+35)/log7(x+6)-1/log7(x-6)<=4
- log2(2*x^2-17*x+35)/log7(x+6)+1/log7(x+6)<=4
- log2(2*x^2-17*x+35)/log7(x-6)-1/log7(x+6)<=4
log2(2*x^2-17*x+35)/log7(x+6)-1/log7(x+6)<=4 inequation
The solution
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[src]
/ 2 \
log\2*x - 17*x + 35/ 1
--------------------- - 1*------------ <= 4
log(x + 6) /log(x + 6)\
log(2)*---------- |----------|
log(7) \ log(7) /
$$\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}} \leq 4$$
log(2*x^2 - 17*x + 35)/(((log(x + 6)/log(7)))*log(2)) - 1/(log(x + 6)/log(7)) <= 4
Detail solution
Given the inequality:
$$\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}} \leq 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}} = 4$$
Solve:
Given the equation
$$\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}} = 4$$
transform
$$\frac{- \log{\left(16 \right)} \log{\left(x + 6 \right)} + \log{\left(7 \right)} \log{\left(2 x^{2} - 17 x + 35 \right)} - \log{\left(7^{\log{\left(2 \right)}} \right)}}{\log{\left(2 \right)} \log{\left(x + 6 \right)}} = 0$$
$$\left(\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}}\right) - 4 = 0$$
Do replacement
$$w = \log{\left(2 x^{2} - 17 x + 35 \right)}$$
Given the equation:
$$\left(\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}}\right) - 4 = 0$$
Use proportions rule:
From a1/b1 = a2/b2 should a1*b2 = a2*b1,
In this case
so we get the equation
$$\frac{\frac{1}{\log{\left(2 \right)}} \log{\left(7 \right)} \log{\left(2 x^{2} - 17 x + 35 \right)}}{4 + \frac{\log{\left(7 \right)}}{\log{\left(x + 6 \right)}}} = 1 \log{\left(x + 6 \right)}$$
$$\frac{\log{\left(7 \right)} \log{\left(2 x^{2} - 17 x + 35 \right)}}{\left(4 + \frac{\log{\left(7 \right)}}{\log{\left(x + 6 \right)}}\right) \log{\left(2 \right)}} = \log{\left(x + 6 \right)}$$
Expand brackets in the left part
Expand brackets in the right part
Looking for similar summands in the left part:
This equation has no roots
do backward replacement
$$\log{\left(2 x^{2} - 17 x + 35 \right)} = w$$
substitute w:
$$x_{1} = 12.1808028470473$$
$$x_{2} = 0.416438278776914$$
$$x_{1} = 12.1808028470473$$
$$x_{2} = 0.416438278776914$$
This roots
$$x_{2} = 0.416438278776914$$
$$x_{1} = 12.1808028470473$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 0.416438278776914$$
=
$$0.316438278776914$$
substitute to the expression
$$\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}} \leq 4$$
$$- \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(0.316438278776914 + 6 \right)}} + \frac{\log{\left(- 0.316438278776914 \cdot 17 + 2 \cdot 0.316438278776914^{2} + 35 \right)}}{\frac{\log{\left(0.316438278776914 + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} \leq 4$$
but
Then
$$x \leq 0.416438278776914$$
no execute
one of the solutions of our inequality is:
$$x \geq 0.416438278776914 \wedge x \leq 12.1808028470473$$
$$\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}} \leq 4$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}} = 4$$
Solve:
Given the equation
$$\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}} = 4$$
transform
$$\frac{- \log{\left(16 \right)} \log{\left(x + 6 \right)} + \log{\left(7 \right)} \log{\left(2 x^{2} - 17 x + 35 \right)} - \log{\left(7^{\log{\left(2 \right)}} \right)}}{\log{\left(2 \right)} \log{\left(x + 6 \right)}} = 0$$
$$\left(\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}}\right) - 4 = 0$$
Do replacement
$$w = \log{\left(2 x^{2} - 17 x + 35 \right)}$$
Given the equation:
$$\left(\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}}\right) - 4 = 0$$
Use proportions rule:
From a1/b1 = a2/b2 should a1*b2 = a2*b1,
In this case
a1 = log(7)*log(35 - 17*x + 2*x^2)/log(2)
b1 = log(6 + x)
a2 = 1
b2 = 1/(4 + log(7)/log(6 + x))
so we get the equation
$$\frac{\frac{1}{\log{\left(2 \right)}} \log{\left(7 \right)} \log{\left(2 x^{2} - 17 x + 35 \right)}}{4 + \frac{\log{\left(7 \right)}}{\log{\left(x + 6 \right)}}} = 1 \log{\left(x + 6 \right)}$$
$$\frac{\log{\left(7 \right)} \log{\left(2 x^{2} - 17 x + 35 \right)}}{\left(4 + \frac{\log{\left(7 \right)}}{\log{\left(x + 6 \right)}}\right) \log{\left(2 \right)}} = \log{\left(x + 6 \right)}$$
Expand brackets in the left part
log7log35+17*x+2*x+24+log+7log6+x)*log2) = log(6 + x)
Expand brackets in the right part
log7log35+17*x+2*x+24+log+7log6+x)*log2) = log6+x
Looking for similar summands in the left part:
log(7)*log(35 - 17*x + 2*x^2)/((4 + log(7)/log(6 + x))*log(2)) = log6+x
This equation has no roots
do backward replacement
$$\log{\left(2 x^{2} - 17 x + 35 \right)} = w$$
substitute w:
$$x_{1} = 12.1808028470473$$
$$x_{2} = 0.416438278776914$$
$$x_{1} = 12.1808028470473$$
$$x_{2} = 0.416438278776914$$
This roots
$$x_{2} = 0.416438278776914$$
$$x_{1} = 12.1808028470473$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} \leq x_{2}$$
For example, let's take the point
$$x_{0} = x_{2} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 0.416438278776914$$
=
$$0.316438278776914$$
substitute to the expression
$$\frac{\log{\left(2 x^{2} - 17 x + 35 \right)}}{\frac{\log{\left(x + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} - 1 \cdot \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(x + 6 \right)}} \leq 4$$
$$- \frac{1}{\frac{1}{\log{\left(7 \right)}} \log{\left(0.316438278776914 + 6 \right)}} + \frac{\log{\left(- 0.316438278776914 \cdot 17 + 2 \cdot 0.316438278776914^{2} + 35 \right)}}{\frac{\log{\left(0.316438278776914 + 6 \right)}}{\log{\left(7 \right)}} \log{\left(2 \right)}} \leq 4$$
1.84206198926356*log(7)
-0.542547825019317*log(7) + ----------------------- <= 4
log(2) but
1.84206198926356*log(7)
-0.542547825019317*log(7) + ----------------------- >= 4
log(2) Then
$$x \leq 0.416438278776914$$
no execute
one of the solutions of our inequality is:
$$x \geq 0.416438278776914 \wedge x \leq 12.1808028470473$$
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x_2 x_1
Solving inequality on a graph