Mister Exam
log1/3x
The solution
You have entered
[src]
log(1)*x
-------- < log(x)*3 - 5/2
3
$$\frac{x \log{\left(1 \right)}}{3} < \log{\left(x \right)} 3 - \frac{5}{2}$$
log(1)*x/3 < log(x)*3 - 1*5/2
Detail solution
Given the inequality:
$$\frac{x \log{\left(1 \right)}}{3} < \log{\left(x \right)} 3 - \frac{5}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x \log{\left(1 \right)}}{3} = \log{\left(x \right)} 3 - \frac{5}{2}$$
Solve:
Given the equation
$$\frac{x \log{\left(1 \right)}}{3} = \log{\left(x \right)} 3 - \frac{5}{2}$$
Transfer the right side of the equation left part with negative sign
$$- 3 \log{\left(x \right)} = - \frac{5}{2}$$
Let's divide both parts of the equation by the multiplier of log =-3
$$\log{\left(x \right)} = \frac{5}{6}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$1 x + 0 = e^{- \frac{5}{\left(-3\right) 2}}$$
simplify
$$x = e^{\frac{5}{6}}$$
$$x_{1} = e^{\frac{5}{6}}$$
$$x_{1} = e^{\frac{5}{6}}$$
This roots
$$x_{1} = e^{\frac{5}{6}}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + e^{\frac{5}{6}}$$
=
$$- \frac{1}{10} + e^{\frac{5}{6}}$$
substitute to the expression
$$\frac{x \log{\left(1 \right)}}{3} < \log{\left(x \right)} 3 - \frac{5}{2}$$
$$\frac{\left(- \frac{1}{10} + e^{\frac{5}{6}}\right) \log{\left(1 \right)}}{3} < \left(-1\right) \frac{5}{2} + \log{\left(- \frac{1}{10} + e^{\frac{5}{6}} \right)} 3$$
5 / 1 5/6\
0 < - - + 3*log|- -- + e |
2 \ 10 /
but
5 / 1 5/6\
0 > - - + 3*log|- -- + e |
2 \ 10 /
Then
$$x < e^{\frac{5}{6}}$$
no execute
the solution of our inequality is:
$$x > e^{\frac{5}{6}}$$
_____
/
-------ο-------
x_1
Solving inequality on a graph
The solution
You have entered
[src]
log(1)*x -------- < log(x)*3 - 5/2 3
$$\frac{x \log{\left(1 \right)}}{3} < \log{\left(x \right)} 3 - \frac{5}{2}$$
log(1)*x/3 < log(x)*3 - 1*5/2
Detail solution
Given the inequality:
$$\frac{x \log{\left(1 \right)}}{3} < \log{\left(x \right)} 3 - \frac{5}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x \log{\left(1 \right)}}{3} = \log{\left(x \right)} 3 - \frac{5}{2}$$
Solve:
Given the equation
$$\frac{x \log{\left(1 \right)}}{3} = \log{\left(x \right)} 3 - \frac{5}{2}$$
Transfer the right side of the equation left part with negative sign
$$- 3 \log{\left(x \right)} = - \frac{5}{2}$$
Let's divide both parts of the equation by the multiplier of log =-3
$$\log{\left(x \right)} = \frac{5}{6}$$
This equation is of the form:
By definition log
then
$$1 x + 0 = e^{- \frac{5}{\left(-3\right) 2}}$$
simplify
$$x = e^{\frac{5}{6}}$$
$$x_{1} = e^{\frac{5}{6}}$$
$$x_{1} = e^{\frac{5}{6}}$$
This roots
$$x_{1} = e^{\frac{5}{6}}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + e^{\frac{5}{6}}$$
=
$$- \frac{1}{10} + e^{\frac{5}{6}}$$
substitute to the expression
$$\frac{x \log{\left(1 \right)}}{3} < \log{\left(x \right)} 3 - \frac{5}{2}$$
$$\frac{\left(- \frac{1}{10} + e^{\frac{5}{6}}\right) \log{\left(1 \right)}}{3} < \left(-1\right) \frac{5}{2} + \log{\left(- \frac{1}{10} + e^{\frac{5}{6}} \right)} 3$$
but
Then
$$x < e^{\frac{5}{6}}$$
no execute
the solution of our inequality is:
$$x > e^{\frac{5}{6}}$$
$$\frac{x \log{\left(1 \right)}}{3} < \log{\left(x \right)} 3 - \frac{5}{2}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{x \log{\left(1 \right)}}{3} = \log{\left(x \right)} 3 - \frac{5}{2}$$
Solve:
Given the equation
$$\frac{x \log{\left(1 \right)}}{3} = \log{\left(x \right)} 3 - \frac{5}{2}$$
Transfer the right side of the equation left part with negative sign
$$- 3 \log{\left(x \right)} = - \frac{5}{2}$$
Let's divide both parts of the equation by the multiplier of log =-3
$$\log{\left(x \right)} = \frac{5}{6}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$1 x + 0 = e^{- \frac{5}{\left(-3\right) 2}}$$
simplify
$$x = e^{\frac{5}{6}}$$
$$x_{1} = e^{\frac{5}{6}}$$
$$x_{1} = e^{\frac{5}{6}}$$
This roots
$$x_{1} = e^{\frac{5}{6}}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + e^{\frac{5}{6}}$$
=
$$- \frac{1}{10} + e^{\frac{5}{6}}$$
substitute to the expression
$$\frac{x \log{\left(1 \right)}}{3} < \log{\left(x \right)} 3 - \frac{5}{2}$$
$$\frac{\left(- \frac{1}{10} + e^{\frac{5}{6}}\right) \log{\left(1 \right)}}{3} < \left(-1\right) \frac{5}{2} + \log{\left(- \frac{1}{10} + e^{\frac{5}{6}} \right)} 3$$
5 / 1 5/6\
0 < - - + 3*log|- -- + e |
2 \ 10 /but
5 / 1 5/6\
0 > - - + 3*log|- -- + e |
2 \ 10 /Then
$$x < e^{\frac{5}{6}}$$
no execute
the solution of our inequality is:
$$x > e^{\frac{5}{6}}$$
_____
/
-------ο-------
x_1
Solving inequality on a graph