What is (0.25)^x>2^{(2x)/(x+1)} ?

The solution to (0.25)^x>2^{(2x)/(x+1)} is x<-2\lor-1<x<0

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(0.25)^x>2^{(2x)/(x+1)}

(0.25)^{x} > 2^{\frac{2 x}{x + 1}}

x < - 2 or - 1 < x < 0

Interval Notation

(- \infty, - 2) \cup (- 1, 0)

Solution steps

(0.25)^{x} > 2^{\frac{2 x}{x + 1}}

Convert decimals to fractions

(\frac{1}{4})^{x} > 2^{\frac{2 x}{x + 1}}

Apply exponent rules

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

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

x < - 2 or - 1 < x < 0

Verify Solutions:

x < - 2 or - 1 < x < 0

The solution is

x < - 2 or - 1 < x < 0

2 x + 2 > 10

8 x - 15 > x^{2}

(\frac{1}{3})^{2}

4 x^{2} - 7 x + 4 = 0

50 x^{2} - 235 x - 75

What is (0.25)^x>2^{(2x)/(x+1)} ?
The solution to (0.25)^x>2^{(2x)/(x+1)} is x<-2\lor-1<x<0