What is log_{x+4}(5x+20)<= log_{x+4}((x+4)^2) ?

The solution to log_{x+4}(5x+20) = 1

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log_{x+4}(5x+20)<= log_{x+4}((x+4)^2)

lo g_{x + 4} (5 x + 20) \leq lo g_{x + 4} ((x + 4)^{2})

- 4 < x < - 3 or x \geq 1

Interval Notation

(- 4, - 3) \cup [1, \infty)

Solution steps

lo g_{x + 4} (5 x + 20) \leq lo g_{x + 4} ((x + 4)^{2})

Apply log rules

- 4 < x < - 3 or x \geq 1

lo g_{x} (x^{2} - 4 x + 4) \geq 0

lo g_{7} (\frac{( 2 x - 6 )}{( 2 x - 1 )}) > 0

(lo g_{0.5} (x))^{2} + lo g_{0.5} (x) - 2 \leq 0

2 - lo g_{10} (4 x) > 0

lo g_{x}^{2} (2 + x) < 1

What is log_{x+4}(5x+20)<= log_{x+4}((x+4)^2) ?
The solution to log_{x+4}(5x+20)<= log_{x+4}((x+4)^2) is -4<x<-3\lor x>= 1