log-1-2-x-2-7x-12-gt-log-x-5-x-2-7x-12-

Question Number 5965 by love math last updated on 07/Jun/16

{l o g}_{\frac{1}{2}} (x^{2} + 7 x + 12) > {l o g}_{x + 5} (x^{2} + 7 x + 12)

Commented by Yozzii last updated on 07/Jun/16

\frac{l n u}{l n 0.5} > \frac{l n u}{l n (x + 5)} (c h a n g e o f b a s e / u = x^{2} + 7 x + 12)

l n u (\frac{1}{l n 0.5} - \frac{1}{l n (x + 5)}) > 0

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(1) l n u > 0 a n d \frac{1}{l n 0.5} > \frac{1}{l n (x + 5)}

u > e^{0} = 1

x^{2} + 7 x + 12 > 1

x^{2} + 7 x + 11 > 0

(x + \frac{7 + \sqrt{5}}{2}) (x + \frac{7 - \sqrt{5}}{2}) > 0

\Rightarrow x \in [(- \infty, - \frac{7 + \sqrt{5}}{2} \approx - 4.62) \cup (\frac{\sqrt{5} - 7}{2} \approx - 2.38, + \infty)] \dots (i)

A l s o, \frac{1}{l n 0.5} > \frac{1}{l n (x + 5)}

\frac{l n (x + 5) - l n 0.5}{(l n 0.5) l n (x + 5)} > 0

\frac{l n (2 (x + 5))}{l n (x + 5)} < 0 (l n 0.5 < 0)

\Rightarrow (1) l n (2 (x + 5)) > 0 & l n (x + 5) < 0

\Rightarrow 2 (x + 5) > 1 & 0 < x + 5 < 1

x > - 4.5 & - 5 < x < - 4 \Rightarrow x \in (- 4.5, - 4) .

\Rightarrow (2) l n (2 (x + 5)) < 0 & l n (x + 5) > 0

\Rightarrow - 5 < x < - 4.5 & x > - 4 (i m p o s s i b l e)

∴ x \in (- 4.5, - 4) \dots . (i i)

T h e r e g i o n o f o v e r l a p o f (i) a n d (i i)

i s e m p t y .

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l n u < 0 a n d \frac{1}{l n 0.5} - \frac{1}{l n (x + 5)} < 0

l n u < 0 \Rightarrow 0 < (x + 4) (x + 3) < 1 \Rightarrow x \in [(- \frac{7 + \sqrt{5}}{2}, - 4) \cup (- 3, \frac{\sqrt{5} - 7}{2})] \dots \dots . . (i)

\frac{1}{l n 0.5} - \frac{1}{l n (x + 5)} < 0 \Rightarrow \frac{l n (2 (x + 5))}{l n (x + 5)} > 0

\Rightarrow (1) l n (2 (x + 5)) > 0 & l n (x + 5) > 0

x > - 4.5 & x > - 4 \Rightarrow x > - 4

(2) l n (2 (x + 5)) < 0 & l n (x + 5) < 0

0 < 2 (x + 5) < 1 & 0 < x + 5 < 1

- 5 < x < - 4.5 & - 5 < x < - 4 \Rightarrow x \in (- 5, - 4.5)

∴ x \in [(- 4, + \infty) \cup (- 5, - 4.5)] \dots . (i i)

R e g i o n o f o v e r l a p o f (i) a n d (i i) i s

i s x \in [(- \frac{5 + \sqrt{7}}{2}, - \frac{9}{2}) \cup (- 3, \frac{\sqrt{5} - 7}{2})] .

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A n s w e r : x \in [(\frac{- 5 - \sqrt{7}}{2}, - \frac{9}{2}) \cup (- 3, \frac{\sqrt{5} - 7}{2})]

Commented by love math last updated on 07/Jun/16

T h e n w h a t n e e d t o d o ?

Commented by prakash jain last updated on 08/Jun/16

Yozzi considered two cases where

\ln (x^{2} + 7 x + 12) > 0

and second case \ln (x^{2} + 7 x + 12) < 0

r a n g e o f v a l i d x i s a l s o i n t h e a n s w e r .

Answered by Ashis last updated on 07/Jun/16

l o g (x + 5) > l o g (1 / 2)

=> l o g (2 x + 10) > 0

=> 2 x + 10 > 1

=> x > - \frac{9}{2}

Commented by Yozzii last updated on 07/Jun/16

W h a t i f x = - 3 ? D o e s t h e i n e q u a l i t y

h o l d i f w e a s s u m e t h e l o g a r i t h m i s r e a l v a l u e d ?

Commented by prakash jain last updated on 08/Jun/16

Just to add to {Yozzi}^{'} s comment .

Complex logarithm and logs of - ve numbers

are complex number and inequality does

not make sense . So real valued logs are

only required .

x \in [- 3, - 4], x^{2} + 7 x + 12 ⩽ 0 so \log cannot be taken .