Mister Exam
log2x+4(x^2+1)<1 inequation
The solution
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/ 2 \ log(2*x) + 4*\x + 1/ < 1
$$4 \left(x^{2} + 1\right) + \log{\left(2 x \right)} < 1$$
4*(x^2 + 1) + log(2*x) < 1
Detail solution
Given the inequality:
$$4 \left(x^{2} + 1\right) + \log{\left(2 x \right)} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$4 \left(x^{2} + 1\right) + \log{\left(2 x \right)} = 1$$
Solve:
$$x_{1} = \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
$$x_{1} = \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
This roots
$$x_{1} = \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{2 \left(e^{1}\right)^{\frac{W\left(\frac{2}{\left(e^{1}\right)^{6}}\right)}{2} + 3}}$$
=
$$- \frac{1}{10} + \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
substitute to the expression
$$4 \left(x^{2} + 1\right) + \log{\left(2 x \right)} < 1$$
$$\log{\left(2 \left(- \frac{1}{10} + \frac{1}{2 \left(e^{1}\right)^{\frac{W\left(\frac{2}{\left(e^{1}\right)^{6}}\right)}{2} + 3}}\right) \right)} + 4 \left(\left(- \frac{1}{10} + \frac{1}{2 \left(e^{1}\right)^{\frac{W\left(\frac{2}{\left(e^{1}\right)^{6}}\right)}{2} + 3}}\right)^{2} + 1\right) < 1$$
Then
$$x < \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
no execute
the solution of our inequality is:
$$x > \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
$$4 \left(x^{2} + 1\right) + \log{\left(2 x \right)} < 1$$
To solve this inequality, we must first solve the corresponding equation:
$$4 \left(x^{2} + 1\right) + \log{\left(2 x \right)} = 1$$
Solve:
$$x_{1} = \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
$$x_{1} = \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
This roots
$$x_{1} = \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
$$x_{0} < x_{1}$$
For example, let's take the point
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{1}{2 \left(e^{1}\right)^{\frac{W\left(\frac{2}{\left(e^{1}\right)^{6}}\right)}{2} + 3}}$$
=
$$- \frac{1}{10} + \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
substitute to the expression
$$4 \left(x^{2} + 1\right) + \log{\left(2 x \right)} < 1$$
$$\log{\left(2 \left(- \frac{1}{10} + \frac{1}{2 \left(e^{1}\right)^{\frac{W\left(\frac{2}{\left(e^{1}\right)^{6}}\right)}{2} + 3}}\right) \right)} + 4 \left(\left(- \frac{1}{10} + \frac{1}{2 \left(e^{1}\right)^{\frac{W\left(\frac{2}{\left(e^{1}\right)^{6}}\right)}{2} + 3}}\right)^{2} + 1\right) < 1$$
2
/ / -6\\
| W\2*e /| / / -6\\
| -3 - --------| | W\2*e /|
| 2 | | -3 - --------| < 1
| 1 e | |1 2 |
4 + 4*|- -- + --------------| + pi*I + log|- - e |
\ 10 2 / \5 /
Then
$$x < \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
no execute
the solution of our inequality is:
$$x > \frac{1}{2 e^{\frac{W\left(\frac{2}{e^{6}}\right)}{2} + 3}}$$
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