Question 1189538
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Solve the following exponential inequalities.
(Cube root of 10)^(1-2x)>0.1^(2x+1)
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<pre>
Your starting inequality is

    {{{(root(3,10))^(1-2x)}}} > {{{0.1^(2x+1)}}}


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

    {{{(root(3,10))^(1-2x)}}} > {{{10^(-(2x+1))}}}.


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

    {{{(1-2x)/3}}} > -(2x+1).


Simplify 

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

    1 - 2x > -6x - 3

    1 + 3 > -6x + 2x

      4   >    -4x

      1   >    -x
  
      x   > -1.


<U>ANSWER</U>.  x > - 1.
</pre>

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.