What is log_{2}(x+1)+log_{x+1}(2)<= 5/2 ?

The solution to log_{2}(x+1)+log_{x+1}(2)<= 5/2 is -1<x<0\lor sqrt(2)-1<= x<= 3

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log_{2}(x+1)+log_{x+1}(2)<= 5/2

lo g_{2} (x + 1) + lo g_{x + 1} (2) \leq \frac{5}{2}

- 1 < x < 0 or 2 - 1 \leq x \leq 3

Interval Notation

(- 1, 0) \cup [2 - 1, 3]

Decimal Notation

- 1 < x < 0 or 0.41421 \dots \leq x \leq 3

Solution steps

lo g_{2} (x + 1) + lo g_{x + 1} (2) \leq \frac{5}{2}

Apply log rules

lo g_{2} (x + 1) + lo g_{2} (2^{\frac{1}{l o g _{2} ( x + 1 )}})

Simplify

lo g_{2} (2^{\frac{1}{l o g _{2} ( x + 1 )}}) : \frac{1}{lo g _{2} ( x + 1 )}

lo g_{2} (x + 1) + \frac{1}{lo g _{2} ( x + 1 )}

Remove parentheses:

(a) = a

lo g_{2} (x + 1) + \frac{1}{lo g _{2} ( x + 1 )}

Let

u = lo g_{2} (x + 1)

u + \frac{1}{u} \leq \frac{5}{2}

u + \frac{1}{u} \leq \frac{5}{2} : u < 0 or \frac{1}{2} \leq u \leq 2

u < 0 or \frac{1}{2} \leq u \leq 2

Substitute back

u = lo g_{2} (x + 1)

lo g_{2} (x + 1) < 0 or \frac{1}{2} \leq lo g_{2} (x + 1) \leq 2

lo g_{2} (x + 1) < 0 : - 1 < x < 0

\frac{1}{2} \leq lo g_{2} (x + 1) \leq 2 : 2 - 1 \leq x \leq 3

Combine the intervals

- 1 < x < 0 or 2 - 1 \leq x \leq 3

\frac{x + 1}{x + 6} > 0

\frac{12 - ( - 2 )}{7} + (- 3) + (2) (- 9)

- 4 ∣ 3 x - 1 ∣ \leq - 16

x^{3} - x < 0

3 + ∣ x - 3 ∣ > \frac{4}{7}

What is log_{2}(x+1)+log_{x+1}(2)<= 5/2 ?
The solution to log_{2}(x+1)+log_{x+1}(2)<= 5/2 is -1<x<0\lor sqrt(2)-1<= x<= 3