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5^(1-x)>0,2 inequation

A inequation with variable

The solution

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 1 - x      
5      > 1/5
51x>155^{1 - x} > \frac{1}{5}
5^(1 - x) > 1/5
Detail solution
Given the inequality:
51x>155^{1 - x} > \frac{1}{5}
To solve this inequality, we must first solve the corresponding equation:
51x=155^{1 - x} = \frac{1}{5}
Solve:
Given the equation:
51x=155^{1 - x} = \frac{1}{5}
or
51x15=05^{1 - x} - \frac{1}{5} = 0
or
55x=155 \cdot 5^{- x} = \frac{1}{5}
or
(15)x=125\left(\frac{1}{5}\right)^{x} = \frac{1}{25}
- this is the simplest exponential equation
Do replacement
v=(15)xv = \left(\frac{1}{5}\right)^{x}
we get
v125=0v - \frac{1}{25} = 0
or
v125=0v - \frac{1}{25} = 0
Move free summands (without v)
from left part to right part, we given:
v=125v = \frac{1}{25}
do backward replacement
(15)x=v\left(\frac{1}{5}\right)^{x} = v
or
x=log(v)log(5)x = - \frac{\log{\left(v \right)}}{\log{\left(5 \right)}}
x1=125x_{1} = \frac{1}{25}
x1=125x_{1} = \frac{1}{25}
This roots
x1=125x_{1} = \frac{1}{25}
is the points with change the sign of the inequality expression.
First define with the sign to the leftmost point:
x0<x1x_{0} < x_{1}
For example, let's take the point
x0=x1110x_{0} = x_{1} - \frac{1}{10}
=
110+125- \frac{1}{10} + \frac{1}{25}
=
350- \frac{3}{50}
substitute to the expression
51x>155^{1 - x} > \frac{1}{5}
51350>155^{1 - - \frac{3}{50}} > \frac{1}{5}
   3/50      
5*5     > 1/5
      

the solution of our inequality is:
x<125x < \frac{1}{25}
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Solving inequality on a graph
012345678-5-4-3-2-1040
Rapid solution [src]
x < 2
x<2x < 2
x < 2
Rapid solution 2 [src]
(-oo, 2)
x in (,2)x\ in\ \left(-\infty, 2\right)
x in Interval.open(-oo, 2)