What is (log_{2}(((x-1))/((x+2))))>0 ?

The solution to (log_{2}(((x-1))/((x+2))))>0 is x<-2

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(log_{2}(((x-1))/((x+2))))>0

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

x < - 2

Interval Notation

(- \infty, - 2)

Solution steps

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

Apply log rules

\frac{x - 1}{x + 2} > 1

\frac{x - 1}{x + 2} > 1 : x < - 2

x < - 2

lo g_{3} (2 x + 5) < 2

(3^{x^{2} - 1} - 2 7^{x + 1}) (lo g_{3} (x + 8) - 2) \leq 0

lo g_{3} (x^{2}) > 2

ln (x) < 1

lo g_{10} (x) (\frac{( 2 x + \frac{2}{5} )}{( 5 ( 1 - x ) )}) > 0

What is (log_{2}(((x-1))/((x+2))))>0 ?
The solution to (log_{2}(((x-1))/((x+2))))>0 is x<-2