What is (0.2)^{((2x-1))/((1-x))}>5 ?

The solution to (0.2)^{((2x-1))/((1-x))}>5 is x 1

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(0.2)^{((2x-1))/((1-x))}>5

(0.2)^{\frac{( 2 x - 1 )}{( 1 - x )}} > 5

x < 0 or x > 1

Interval Notation

(- \infty, 0) \cup (1, \infty)

Solution steps

(0.2)^{\frac{( 2 x - 1 )}{( 1 - x )}} > 5

Apply exponent rules

- \frac{2 x - 1}{1 - x} > 1

- \frac{2 x - 1}{1 - x} > 1 : x < 0 or x > 1

x < 0 or x > 1

Verify Solutions:

x < 0 or x > 1

The solution is

x < 0 or x > 1

4^{x} < 2^{x + 1} + 3

3^{x} + 3^{2 - x} > 10

\frac{[ 2 ^{1 - x} - 2 ^{x} + 1 ]}{[ 2 ^{x} - 1 ]} < 0

e^{x - 1} > 0

e^{- x} < 1

What is (0.2)^{((2x-1))/((1-x))}>5 ?
The solution to (0.2)^{((2x-1))/((1-x))}>5 is x<0\lor x>1