What is log_{2}(x^2-6x)<= 3+log_{2}(1-x) ?

The solution to log_{2}(x^2-6x)<= 3+log_{2}(1-x) is -4<= x<0

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log_{2}(x^2-6x)<= 3+log_{2}(1-x)

lo g_{2} (x^{2} - 6 x) \leq 3 + lo g_{2} (1 - x)

- 4 \leq x < 0

Interval Notation

[- 4, 0)

Solution steps

lo g_{2} (x^{2} - 6 x) \leq 3 + lo g_{2} (1 - x)

Apply log rules

x^{2} - 6 x \leq 8 (1 - x) and x^{2} - 6 x > 0

x^{2} - 6 x \leq 8 (1 - x) and x^{2} - 6 x > 0 : - 4 \leq x < 0

- 4 \leq x < 0

Find positive values for logs:

x < 0

Combine the intervals

- 4 \leq x < 0 and x < 0

Merge Overlapping Intervals

- 4 \leq x < 0

2 x^{2} + 5 x - 9 = 0

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

- \frac{π}{2} - \frac{π}{4}

5 x^{2} = x

n^{2} - 12 n + 11

What is log_{2}(x^2-6x)<= 3+log_{2}(1-x) ?
The solution to log_{2}(x^2-6x)<= 3+log_{2}(1-x) is -4<= x<0