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Solving inequalities/ lgx>=1

lgx>=1 inequation

A inequation with variable

The solution

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log(x) >= 1
log⁡(x)≥1\log{\left(x \right)} \geq 1 
log(x) >= 1
Detail solution
Given the inequality:
log⁡(x)≥1\log{\left(x \right)} \geq 1
To solve this inequality, we must first solve the corresponding equation:
log⁡(x)=1\log{\left(x \right)} = 1
Solve:
Given the equation
log⁡(x)=1\log{\left(x \right)} = 1
log⁡(x)=1\log{\left(x \right)} = 1
This equation is of the form:
log(v)=p

By definition log
v=e^p

then
x=e1−1x = e^{1^{-1}}
simplify
x=ex = e
x1=ex_{1} = e
x1=ex_{1} = e
This roots
x1=ex_{1} = e
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
x0≤x1x_{0} \leq x_{1}
For example, let's take the point
x0=x1−110x_{0} = x_{1} - \frac{1}{10}
=
−110+e- \frac{1}{10} + e
=
−110+e- \frac{1}{10} + e
substitute to the expression
log⁡(x)≥1\log{\left(x \right)} \geq 1
log⁡(−110+e)≥1\log{\left(- \frac{1}{10} + e \right)} \geq 1
log(-1/10 + E) >= 1

but
log(-1/10 + E) < 1

Then
x≤ex \leq e
no execute
the solution of our inequality is:
x≥ex \geq e
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Solving inequality on a graph
02468-4-210-1010
Rapid solution [src] 
E <= x
e≤xe \leq x 
E <= x
Rapid solution 2 [src] 
[E, oo)
x in [e,∞)x\ in\ \left[e, \infty\right) 
x in Interval(E, oo)
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