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)   (Show Source): You can put this solution on YOUR website!

............take log of both sides

.......note that

......



.......,











Answer by ikleyn(52887)   (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

     > 


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

     > .


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

     > -(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.

=================

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.



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