SOLUTION: Solve the following exponential inequalities. (Cube root of 10)^(1-2x)>0.1^(2x+1)

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Question 1189538: Solve the following exponential inequalities.
(Cube root of 10)^(1-2x)>0.1^(2x+1)

Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!

%28root%283%2C10%29%29%5E%281-2x%29+%3E+0.1%5E%282x%2B1%29............take log of both sides

log%28%28root%283%2C10%29%29%5E%281-2x%29+%29%3E+log%280.1%5E%282x%2B1%29%29.......note that root%283%2C10%29=10%5E%281%2F3%29

%281-2x%29+%2Alog%28%2810%5E%281%2F3%29%29%29%3E+%282x%2B1%29%2Alog%280.1%29......0.1=1%2F10

%281-2x%29%2F+%282x%2B1%29%3E+log%281%2F10%29%2Flog%28%2810%5E%281%2F3%29%29%29

.......log%281%29=0, log%2810%29=1

%281-2x%29%2F+%282x%2B1%29%3E+%280-1%29%2F%28%281%2F3%29%2A1%29

%281-2x%29%2F+%282x%2B1%29%3E+-1%2F%281%2F3%29
%281-2x%29%2F+%282x%2B1%29%3E+-3
1-2x%3E+-3%282x%2B1%29
1-2x%3E+-6x-3
6x-2x%3E+-1-3
4x%3E+-4
x%3E-1

Answer by ikleyn(52887) About Me  (Show Source):
You can put this solution on YOUR website!
.
Solve the following exponential inequalities.
(Cube root of 10)^(1-2x)>0.1^(2x+1)
~~~~~~~~~~~~~~~~ 

Your starting inequality is

    %28root%283%2C10%29%29%5E%281-2x%29 > 0.1%5E%282x%2B1%29


Write right side with the base 10 to have the same base in both sides of the inequality

    %28root%283%2C10%29%29%5E%281-2x%29 > 10%5E%28-%282x%2B1%29%29.


Re-write it equivalently in this form

    10^((1-2x)/3) > 10^(-(2x+1)).


Exponential function in both sides is monotonically increasing - THEREFORE, from the last inequality you have

    %281-2x%29%2F3 > -(2x+1).


Simplify 

    1 - 2x > -3*(2x+1)

    1 - 2x > -6x - 3

    1 + 3 > -6x + 2x

      4   >    -4x

      1   >    -x
  
      x   > -1.


ANSWER.  x > - 1.

Solved.

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May god save you from solving this inequality in a way as @MathLover1 does it.

This woman can not solve equally as she can not teach.