Mister Exam
x^2*log16(x)>=log16(x)^5+(x)*log2(x) inequation
The solution
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2 5 x *log(x) / log(x)\ x*log(x) --------- >= |-------| + -------- log(16) \log(16)/ log(2)
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(16 \right)}} \geq \left(\frac{\log{\left(x \right)}}{\log{\left(16 \right)}}\right)^{5} + \frac{x \log{\left(x \right)}}{\log{\left(2 \right)}}$$
x^2*log(x)/log(16) >= x*log(x)/log(2) + (log(x)/log(16))^5
Detail solution
Given the inequality:
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(16 \right)}} \geq \left(\frac{\log{\left(x \right)}}{\log{\left(16 \right)}}\right)^{5} + \frac{x \log{\left(x \right)}}{\log{\left(2 \right)}}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(16 \right)}} = \left(\frac{\log{\left(x \right)}}{\log{\left(16 \right)}}\right)^{5} + \frac{x \log{\left(x \right)}}{\log{\left(2 \right)}}$$
Solve:
$$x_{1} = 1$$
$$x_{2} = 4.01574087821673$$
$$x_{1} = 1$$
$$x_{2} = 4.01574087821673$$
This roots
$$x_{1} = 1$$
$$x_{2} = 4.01574087821673$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$0.9$$
substitute to the expression
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(16 \right)}} \geq \left(\frac{\log{\left(x \right)}}{\log{\left(16 \right)}}\right)^{5} + \frac{x \log{\left(x \right)}}{\log{\left(2 \right)}}$$
$$\frac{0.9^{2} \log{\left(0.9 \right)}}{\log{\left(16 \right)}} \geq \frac{0.9 \log{\left(0.9 \right)}}{\log{\left(2 \right)}} + \left(\frac{\log{\left(0.9 \right)}}{\log{\left(16 \right)}}\right)^{5}$$
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:
$$x \leq 1$$
$$x \geq 4.01574087821673$$
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(16 \right)}} \geq \left(\frac{\log{\left(x \right)}}{\log{\left(16 \right)}}\right)^{5} + \frac{x \log{\left(x \right)}}{\log{\left(2 \right)}}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(16 \right)}} = \left(\frac{\log{\left(x \right)}}{\log{\left(16 \right)}}\right)^{5} + \frac{x \log{\left(x \right)}}{\log{\left(2 \right)}}$$
Solve:
$$x_{1} = 1$$
$$x_{2} = 4.01574087821673$$
$$x_{1} = 1$$
$$x_{2} = 4.01574087821673$$
This roots
$$x_{1} = 1$$
$$x_{2} = 4.01574087821673$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 1$$
=
$$0.9$$
substitute to the expression
$$\frac{x^{2} \log{\left(x \right)}}{\log{\left(16 \right)}} \geq \left(\frac{\log{\left(x \right)}}{\log{\left(16 \right)}}\right)^{5} + \frac{x \log{\left(x \right)}}{\log{\left(2 \right)}}$$
$$\frac{0.9^{2} \log{\left(0.9 \right)}}{\log{\left(16 \right)}} \geq \frac{0.9 \log{\left(0.9 \right)}}{\log{\left(2 \right)}} + \left(\frac{\log{\left(0.9 \right)}}{\log{\left(16 \right)}}\right)^{5}$$
0.0948244640920437 1.29834299005241e-5
-0.0853420176828393 - ------------------ - -------------------
------------------- >= log(2) 5
log(16) log (16)
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:
$$x \leq 1$$
$$x \geq 4.01574087821673$$
Solving inequality on a graph