log35(35x+2)<=1 inequation

Given the inequality:
$$\frac{\log{\left(35 x + 2 \right)}}{\log{\left(35 \right)}} \leq 1$$
To solve this inequality, we must first solve the corresponding equation:
$$\frac{\log{\left(35 x + 2 \right)}}{\log{\left(35 \right)}} = 1$$
Solve:
Given the equation
$$\frac{\log{\left(35 x + 2 \right)}}{\log{\left(35 \right)}} = 1$$
$$\frac{\log{\left(35 x + 2 \right)}}{\log{\left(35 \right)}} = 1$$
Let's divide both parts of the equation by the multiplier of log =1/log(35)
$$\log{\left(35 x + 2 \right)} = \log{\left(35 \right)}$$
This equation is of the form:

log(v)=p

By definition log

v=e^p

then
$$35 x + 2 = e^{\frac{1}{\frac{1}{\log{\left(35 \right)}}}}$$
simplify
$$35 x + 2 = 35$$
$$35 x = 33$$
$$x = \frac{33}{35}$$
$$x_{1} = \frac{33}{35}$$
$$x_{1} = \frac{33}{35}$$
This roots
$$x_{1} = \frac{33}{35}$$
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
$$x_{0} = x_{1} - \frac{1}{10}$$
=
$$- \frac{1}{10} + \frac{33}{35}$$
=
$$\frac{59}{70}$$
substitute to the expression
$$\frac{\log{\left(35 x + 2 \right)}}{\log{\left(35 \right)}} \leq 1$$
$$\frac{\log{\left(2 + 35 \cdot \frac{59}{70} \right)}}{\log{\left(35 \right)}} \leq 1$$

log(63/2)     
--------- <= 1
 log(35)      

the solution of our inequality is:
$$x \leq \frac{33}{35}$$

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