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

The solution to log_{6}(x-3)+log_{6}(x+2)<2 is 3<x<7

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

lo g_{6} (x - 3) + lo g_{6} (x + 2) < 2

3 < x < 7

Interval Notation

(3, 7)

Solution steps

lo g_{6} (x - 3) + lo g_{6} (x + 2) < 2

Apply log rules

(x - 3) (x + 2) < 36 and (x - 3) (x + 2) > 0

(x - 3) (x + 2) < 36 and (x - 3) (x + 2) > 0 : - 6 < x < - 2 or 3 < x < 7

- 6 < x < - 2 or 3 < x < 7

Find positive values for logs:

x > 3

Combine the intervals

(- 6 < x < - 2 or 3 < x < 7) and x > 3

Merge Overlapping Intervals

3 < x < 7

s = 3.5 (\frac{3 π}{2})

x^{6} + 8

\frac{x + 4}{5} - x + 4 = \frac{x}{3} \frac{x - 2}{2}

lo g_{3} (x) \geq - 4

\frac{4}{5} + \frac{2}{4}

What is log_{6}(x-3)+log_{6}(x+2)<2 ?
The solution to log_{6}(x-3)+log_{6}(x+2)<2 is 3<x<7