Log 3 x-72 a2-3-0- - a - R- if x lies in the interval -a-b then 3 a-b is

By definition of log_a x, we must have 2a^2 + 3 gt; 0 and 3x+7 gt;0 1st is always true and 2nd ⇒ x gt; 7/3 Now log_3x+7 (2a^2 + 3) lt; 0 ⇒log (2a^2 ...

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Standard IX

Mathematics

Logarithmic Inequalities

log 3 x+72 a2...

Question

$l o g_{3 x + 7} (2 a^{2} + 3) < 0, \forall a ϵ R$ , if x lies in the interval (-a,-b) then 3a+ b is ___

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Solution

By definition of $l o g_{a} x$ , we must have
$2 a^{2} + 3 > 0$ and $3 x + 7 > 0$
1st is always true and 2nd $\Rightarrow x > - \frac{7}{3}$
Now $l o g_{3 x + 7} (2 a^{2} + 3) < 0$
$\Rightarrow \frac{l o g (2 a^{2} + 3)}{l o g (3 x + 7)} < 0$
Since $2 a^{2} + 3 > 1$
$∴ l o g (2 a^{2} + 3) > l o g 1 = 0$ i.e. +ive
Hence $l o g (3 x + 7)$ must be - ive
$∴ 3 x + 7 < 1 \Rightarrow x < - 2 \Rightarrow x ϵ (- \frac{7}{3}, - 2)$

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log3x+7(2a2+3)<0,∀ aϵR, if x lies in the interval (-a,-b) then 3a+ b is ___

By definition of loga x, we must have 2a2+3>0 and 3x+7>0 1st is always true and 2nd ⇒x>−73 Now log3x+7(2a2+3)<0 ⇒log (2a2+3)log (3x+7)<0 Since 2a2+3>1 ∴log(2a2+3)>log 1=0 i.e. +ive Hence log(3x+7) must be - ive ∴3x+7<1⇒x<−2⇒xϵ(−73,−2)

$l o g_{3 x + 7} (2 a^{2} + 3) < 0, \forall a ϵ R$ , if x lies in the interval (-a,-b) then 3a+ b is ___