Mister Exam
log2(16-2x)>3 inequation
The solution
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[src]
log(16 - 2*x)
------------- > 3
log(2)
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} > 3$$
log(16 - 2*x)/log(2) > 3
Detail solution
Given the inequality:
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} > 3$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} = 3$$
Solve:
Given the equation
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} = 3$$
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} = 3$$
Let's divide both parts of the equation by the multiplier of log =1/log(2)
$$\log{\left(16 - 2 x \right)} = 3 \log{\left(2 \right)}$$
This equation is of the form:
By definition log
then
$$16 - 2 x = e^{\frac{3}{\frac{1}{\log{\left(2 \right)}}}}$$
simplify
$$16 - 2 x = 8$$
$$- 2 x = -8$$
$$x = 4$$
$$x_{1} = 4$$
$$x_{1} = 4$$
This roots
$$x_{1} = 4$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 4$$
=
$$\frac{39}{10}$$
substitute to the expression
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} > 3$$
$$\frac{\log{\left(16 - \frac{2 \cdot 39}{10} \right)}}{\log{\left(2 \right)}} > 3$$
the solution of our inequality is:
$$x < 4$$
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} > 3$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} = 3$$
Solve:
Given the equation
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} = 3$$
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} = 3$$
Let's divide both parts of the equation by the multiplier of log =1/log(2)
$$\log{\left(16 - 2 x \right)} = 3 \log{\left(2 \right)}$$
This equation is of the form:
log(v)=p
By definition log
v=e^p
then
$$16 - 2 x = e^{\frac{3}{\frac{1}{\log{\left(2 \right)}}}}$$
simplify
$$16 - 2 x = 8$$
$$- 2 x = -8$$
$$x = 4$$
$$x_{1} = 4$$
$$x_{1} = 4$$
This roots
$$x_{1} = 4$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + 4$$
=
$$\frac{39}{10}$$
substitute to the expression
$$\frac{\log{\left(16 - 2 x \right)}}{\log{\left(2 \right)}} > 3$$
$$\frac{\log{\left(16 - \frac{2 \cdot 39}{10} \right)}}{\log{\left(2 \right)}} > 3$$
log(41/5) --------- > 3 log(2)
the solution of our inequality is:
$$x < 4$$
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