What is solvefor q,q13log_{x^2}(2+x)<1 ?

The solution to solvefor q,q13log_{x^2}(2+x) 1

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solvefor q,q13log_{x^2}(2+x)<1

solve for

q, q 13 lo g_{x^{2}} (2 + x) < 1

q < \frac{2}{13 lo g _{x} ( 2 + x )}; x > 1

Solution steps

q \cdot 13 lo g_{x^{2}} (2 + x) < 1

Simplify

13 lo g_{x^{2}} (2 + x) : \frac{13}{2} lo g_{x} (2 + x)

q \frac{13}{2} lo g_{x} (2 + x) < 1

Multiply both sides by

2

13 q lo g_{x} (2 + x) < 2

Divide both sides by

13 lo g_{x} (2 + x); x > 1

q < \frac{2}{13 lo g _{x} ( 2 + x )}; x > 1

4 > 11 - 3 x

3 \frac{( x + 1 )}{2} - x - \frac{1}{4} < \frac{1}{2}

\frac{( 4 x + 2 )}{3} \leq \frac{( 25 - 2 x )}{4}

4 (x + 7) > 3 (x + 2)

5 - 2 x \leq 3

What is solvefor q,q13log_{x^2}(2+x)<1 ?
The solution to solvefor q,q13log_{x^2}(2+x)<1 is q< 2/(13log_{x)(2+x)};\quad x>1