What is 5^{2x}-5^{x+1}+5>= 5^{x+1}-20 ?

The solution to 5^{2x}-5^{x+1}+5>= 5^{x+1}-20 is True for all x

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5^{2x}-5^{x+1}+5>= 5^{x+1}-20

5^{2 x} - 5^{x + 1} + 5 \geq 5^{x + 1} - 20

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

5^{2 x} - 5^{x + 1} + 5 \geq 5^{x + 1} - 20

Apply exponent rules

(5^{x})^{2} - 5^{x} \cdot 5 + 5 \geq 5^{x} \cdot 5 - 20

Let

u = 5^{x}

u^{2} - u \cdot 5 + 5 \geq u \cdot 5 - 20

u^{2} - u \cdot 5 + 5 \geq u \cdot 5 - 20 :

True for all

u

True for all u

u = 5^{x},

therefore

True for all x

4 x^{2} - 4 x = - 9

100 v^{2} - 182 v + 82.56 = 0

\frac{x ^{\frac{2}{3}}}{x ^{2}}

x^{3} (y^{- 5}) (x^{- 8})

∣ x + 2 ∣ < ∣ 2 x ∣

What is 5^{2x}-5^{x+1}+5>= 5^{x+1}-20 ?
The solution to 5^{2x}-5^{x+1}+5>= 5^{x+1}-20 is True for all x