Mister Exam
log3(x+2)<=1 inequation
The solution
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log(x + 2) ---------- <= 1 log(3)
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} \leq 1$$
log(x + 2)/log(3) <= 1
Detail solution
Given the inequality:
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} \leq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} = 1$$
Solve:
Given the equation
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} = 1$$
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} = 1$$
Let's divide both parts of the equation by the multiplier of log =1/log(3)
$$\log{\left(x + 2 \right)} = \log{\left(3 \right)}$$
This equation is of the form:
By definition log
then
$$x + 2 = e^{\frac{1}{\frac{1}{\log{\left(3 \right)}}}}$$
simplify
$$x + 2 = 3$$
$$x = 1$$
$$x_{1} = 1$$
$$x_{1} = 1$$
This roots
$$x_{1} = 1$$
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
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} \leq 1$$
$$\frac{\log{\left(\frac{9}{10} + 2 \right)}}{\log{\left(3 \right)}} \leq 1$$
the solution of our inequality is:
$$x \leq 1$$
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} \leq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} = 1$$
Solve:
Given the equation
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} = 1$$
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} = 1$$
Let's divide both parts of the equation by the multiplier of log =1/log(3)
$$\log{\left(x + 2 \right)} = \log{\left(3 \right)}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$x + 2 = e^{\frac{1}{\frac{1}{\log{\left(3 \right)}}}}$$
simplify
$$x + 2 = 3$$
$$x = 1$$
$$x_{1} = 1$$
$$x_{1} = 1$$
This roots
$$x_{1} = 1$$
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
$$\frac{\log{\left(x + 2 \right)}}{\log{\left(3 \right)}} \leq 1$$
$$\frac{\log{\left(\frac{9}{10} + 2 \right)}}{\log{\left(3 \right)}} \leq 1$$
/29\ log|--| \10/ <= 1 ------- log(3)
the solution of our inequality is:
$$x \leq 1$$
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Solving inequality on a graph