What is log_{2}(1+x^{-1})+log_{1/2}((1+x)/4)>= 1 ?

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

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log_{2}(1+x^{-1})+log_{1/2}((1+x)/4)>= 1

lo g_{2} (1 + x^{- 1}) + lo g_{\frac{1}{2}} (\frac{1 + x}{4}) \geq 1

0 < x \leq 2

Interval Notation

(0, 2]

Solution steps

lo g_{2} (1 + x^{- 1}) + lo g_{\frac{1}{2}} (\frac{1 + x}{4}) \geq 1

Apply log rules

\frac{4 ( x + 1 )}{x ( 1 + x )} \geq 2

\frac{4 ( x + 1 )}{x ( 1 + x )} \geq 2 : 0 < x \leq 2

0 < x \leq 2

Find positive values for logs:

x > 0

Combine the intervals

0 < x \leq 2 and x > 0

Merge Overlapping Intervals

0 < x \leq 2

x - ln (x) > 0

2 \geq 1 + ln (b)

lo g_{6} (x^{2}) \geq lo g_{6} (9)

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

lo g_{x} (\frac{2 x - 3}{4}) > 2

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