Mister Exam
log2(4x-7)≤log2(x+1) inequation
The solution
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log(4*x - 7) log(x + 1) ------------ <= ---------- log(2) log(2)
$$\frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} \leq \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}}$$
log(4*x - 1*7)/log(2) <= log(x + 1)/log(2)
Detail solution
Given the inequality:
$$\frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} \leq \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} = \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}}$$
Solve:
Given the equation
$$\frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} = \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}}$$
transform
$$\frac{- \log{\left(x + 1 \right)} + \log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} = 0$$
$$- \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}} + \frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} = 0$$
Do replacement
$$w = \log{\left(2 \right)}$$
Given the equation:
$$- \frac{\log{\left(x + 1 \right)}}{w} + \frac{\log{\left(4 x - 7 \right)}}{w} = 0$$
Use proportions rule:
From a1/b1 = a2/b2 should a1*b2 = a2*b1,
In this case
so we get the equation
$$w \log{\left(4 x - 7 \right)} = w \log{\left(x + 1 \right)}$$
$$w \log{\left(4 x - 7 \right)} = w \log{\left(x + 1 \right)}$$
Expand brackets in the left part
Expand brackets in the right part
Move free summands (without w)
from left part to right part, we given:
$$w \log{\left(4 x - 7 \right)} + 7 = w \log{\left(x + 1 \right)} + 7$$
Divide both parts of the equation by (7 + w*log(-7 + 4*x))/w
We get the answer: w = 0
do backward replacement
$$\log{\left(2 \right)} = w$$
substitute w:
$$x_{1} = \frac{8}{3}$$
$$x_{1} = \frac{8}{3}$$
This roots
$$x_{1} = \frac{8}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{8}{3}$$
=
$$\frac{77}{30}$$
substitute to the expression
$$\frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} \leq \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}}$$
$$\frac{\log{\left(\left(-1\right) 7 + 4 \cdot \frac{77}{30} \right)}}{\log{\left(2 \right)}} \leq \frac{\log{\left(1 + \frac{77}{30} \right)}}{\log{\left(2 \right)}}$$
the solution of our inequality is:
$$x \leq \frac{8}{3}$$
$$\frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} \leq \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}}$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} = \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}}$$
Solve:
Given the equation
$$\frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} = \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}}$$
transform
$$\frac{- \log{\left(x + 1 \right)} + \log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} = 0$$
$$- \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}} + \frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} = 0$$
Do replacement
$$w = \log{\left(2 \right)}$$
Given the equation:
$$- \frac{\log{\left(x + 1 \right)}}{w} + \frac{\log{\left(4 x - 7 \right)}}{w} = 0$$
Use proportions rule:
From a1/b1 = a2/b2 should a1*b2 = a2*b1,
In this case
a1 = log(-7 + 4*x)
b1 = w
a2 = log(1 + x)
b2 = w
so we get the equation
$$w \log{\left(4 x - 7 \right)} = w \log{\left(x + 1 \right)}$$
$$w \log{\left(4 x - 7 \right)} = w \log{\left(x + 1 \right)}$$
Expand brackets in the left part
w*log-7+4*x = w*log(1 + x)
Expand brackets in the right part
w*log-7+4*x = w*log1+x
Move free summands (without w)
from left part to right part, we given:
$$w \log{\left(4 x - 7 \right)} + 7 = w \log{\left(x + 1 \right)} + 7$$
Divide both parts of the equation by (7 + w*log(-7 + 4*x))/w
w = 7 + w*log(1 + x) / ((7 + w*log(-7 + 4*x))/w)
We get the answer: w = 0
do backward replacement
$$\log{\left(2 \right)} = w$$
substitute w:
$$x_{1} = \frac{8}{3}$$
$$x_{1} = \frac{8}{3}$$
This roots
$$x_{1} = \frac{8}{3}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{8}{3}$$
=
$$\frac{77}{30}$$
substitute to the expression
$$\frac{\log{\left(4 x - 7 \right)}}{\log{\left(2 \right)}} \leq \frac{\log{\left(x + 1 \right)}}{\log{\left(2 \right)}}$$
$$\frac{\log{\left(\left(-1\right) 7 + 4 \cdot \frac{77}{30} \right)}}{\log{\left(2 \right)}} \leq \frac{\log{\left(1 + \frac{77}{30} \right)}}{\log{\left(2 \right)}}$$
/49\ /107\ log|--| log|---| \15/ <= \ 30/ ------- -------- log(2) log(2)
the solution of our inequality is:
$$x \leq \frac{8}{3}$$
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