What is log_{1/x}(2.5x-1)>=-2 ?

The solution to log_{1/x}(2.5x-1)>=-2 is 0.5 = sqrt(0.5625)+1.25

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log_{1/x}(2.5x-1)>=-2

lo g_{\frac{1}{x}} (2.5 x - 1) \geq - 2

0.5 \leq x < 1 or x \geq 0.5625 + 1.25

Interval Notation

[0.5, 1) \cup [0.5625 + 1.25, \infty)

Decimal Notation

0.5 \leq x < 1 or x \geq 2

Solution steps

lo g_{\frac{1}{x}} (2.5 x - 1) \geq - 2

Apply log rules

- 0.5625 + 1.25 \leq x < 1 or x \geq 0.5625 + 1.25

- 0.5625 + 1.25 \leq x < 1 or x \geq 0.5625 + 1.25 : 0.5 \leq x < 1 or x \geq 0.5625 + 1.25

0.5 \leq x < 1 or x \geq 0.5625 + 1.25

2 x + 23 = 6 x + 47

2^{k + 1} + 2^{k + 1}

s^{2} - 7 s = 0

- \frac{1}{2} + 9

\frac{x ^{- 6}}{x ^{- 2}}

What is log_{1/x}(2.5x-1)>=-2 ?
The solution to log_{1/x}(2.5x-1)>=-2 is 0.5<= x<1\lor x>= sqrt(0.5625)+1.25