Mister Exam
x^2*log25(x)≥log25(x^3)+x*log5(x) inequation
The solution
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2 / 3\ x *log(x) log\x / x*log(x) --------- >= ------- + -------- log(25) log(25) log(5)
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} \geq \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} + \frac{\log{\left(x^{3} \right)}}{\log{\left(25 \right)}}$$
x^2*log(x)/log(25) >= x*log(x)/log(5) + log(x^3)/log(25)
Detail solution
Given the inequality:
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} \geq \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} + \frac{\log{\left(x^{3} \right)}}{\log{\left(25 \right)}}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} = \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} + \frac{\log{\left(x^{3} \right)}}{\log{\left(25 \right)}}$$
Solve:
Given the equation
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} = \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} + \frac{\log{\left(x^{3} \right)}}{\log{\left(25 \right)}}$$
transform
$$\frac{\left(x^{2} + 2 x - 3\right) \log{\left(x \right)}}{2 \log{\left(5 \right)}} = 0$$
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} + \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} - \frac{3 \log{\left(x \right)}}{\log{\left(25 \right)}} = 0$$
Do replacement
$$w = \log{\left(x \right)}$$
Expand brackets in the left part
Divide both parts of the equation by (-3*w/log(25) + w*x/log(5) + w*x^2/log(25))/w
We get the answer: w = 0
do backward replacement
$$\log{\left(x \right)} = w$$
Given the equation
$$\log{\left(x \right)} = w$$
$$\log{\left(x \right)} = w$$
This equation is of the form:
By definition log
then
$$1 x + 0 = e^{\frac{w}{1}}$$
simplify
$$x = e^{w}$$
substitute w:
$$x_{1} = 1$$
$$x_{2} = 3$$
$$x_{1} = 1$$
$$x_{2} = 3$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} \geq \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} + \frac{\log{\left(x^{3} \right)}}{\log{\left(25 \right)}}$$
$$\frac{\left(\frac{9}{10}\right)^{2} \log{\left(\frac{9}{10} \right)}}{\log{\left(25 \right)}} \geq \frac{\log{\left(\left(\frac{9}{10}\right)^{3} \right)}}{\log{\left(25 \right)}} + \frac{9 \log{\left(\frac{9}{10} \right)}}{10 \log{\left(5 \right)}}$$
one of the solutions of our inequality is:
$$x \leq 1$$
Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq 1$$
$$x \geq 3$$
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} \geq \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} + \frac{\log{\left(x^{3} \right)}}{\log{\left(25 \right)}}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} = \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} + \frac{\log{\left(x^{3} \right)}}{\log{\left(25 \right)}}$$
Solve:
Given the equation
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} = \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} + \frac{\log{\left(x^{3} \right)}}{\log{\left(25 \right)}}$$
transform
$$\frac{\left(x^{2} + 2 x - 3\right) \log{\left(x \right)}}{2 \log{\left(5 \right)}} = 0$$
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} + \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} - \frac{3 \log{\left(x \right)}}{\log{\left(25 \right)}} = 0$$
Do replacement
$$w = \log{\left(x \right)}$$
Expand brackets in the left part
-3*w/log25 + w*x/log5 + w*x^2/log25 = 0
Divide both parts of the equation by (-3*w/log(25) + w*x/log(5) + w*x^2/log(25))/w
w = 0 / ((-3*w/log(25) + w*x/log(5) + w*x^2/log(25))/w)
We get the answer: w = 0
do backward replacement
$$\log{\left(x \right)} = w$$
Given the equation
$$\log{\left(x \right)} = w$$
$$\log{\left(x \right)} = w$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$1 x + 0 = e^{\frac{w}{1}}$$
simplify
$$x = e^{w}$$
substitute w:
$$x_{1} = 1$$
$$x_{2} = 3$$
$$x_{1} = 1$$
$$x_{2} = 3$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$\frac{9}{10}$$
substitute to the expression
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(25 \right)}} \geq \frac{x \log{\left(x \right)}}{\log{\left(5 \right)}} + \frac{\log{\left(x^{3} \right)}}{\log{\left(25 \right)}}$$
$$\frac{\left(\frac{9}{10}\right)^{2} \log{\left(\frac{9}{10} \right)}}{\log{\left(25 \right)}} \geq \frac{\log{\left(\left(\frac{9}{10}\right)^{3} \right)}}{\log{\left(25 \right)}} + \frac{9 \log{\left(\frac{9}{10} \right)}}{10 \log{\left(5 \right)}}$$
/729 \
81*log(9/10) log|----|
------------ >= \1000/ 9*log(9/10)
100*log(25) --------- + -----------
log(25) 10*log(5) one of the solutions of our inequality is:
$$x \leq 1$$
_____ _____
\ /
-------•-------•-------
x_1 x_2Other solutions will get with the changeover to the next point
etc.
The answer:
$$x \leq 1$$
$$x \geq 3$$
Solving inequality on a graph