Find x: 48log_a 4 + 5log


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Question Number 41797 by Tawa1 last updated on 12/Aug/18

$Find x : {48 \log}_{a} 4 + {5 \log}_{4} a = \frac{a}{8}$
	Answered by alex041103 last updated on 13/Aug/18

	${l o g}_{b} a = \frac{l n a}{l n b} a n d {l o g}_{a} b = \frac{l n b}{l n a}$ $\Rightarrow {l o g}_{4} a = \frac{1}{{l o g}_{a} 4}$ $\Rightarrow l e t {l o g}_{4} a = t$ $\Rightarrow 48 t^{2} + 5 - \frac{a}{8} t = 0$ $\Rightarrow t_{1, 2} = \frac{\frac{a}{8} \pm \sqrt{\frac{a^{2}}{64} - 960}}{96} = {l o g}_{4} a$ $. . . . W f u n c t i o n . . .$
	Commented by Tawa1 last updated on 13/Aug/18

	$God bless you sir$
	Commented by MrW3 last updated on 13/Aug/18

	$w o r k i n g i s n o t c o r r e c t s i r!$ $s o l u t i o n f o r$ ${a x}^{2} + b x + c = 0$ $i s x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a} o n l y$ $w h e n a, b, c a r e c o n s t a n t s, i . e . w h e n$ $t h e y a r e i n d e p e n t e n t f r o m x!$ $b u t i n 48 t^{2} + 5 - \frac{a}{8} t = 0 t a n d a a r e d e p e n d e n t$ $f r o m e a c h o t h e r, s o y o u m a y n o t a p p l y$ $t h e f o r m u l a f o r q u a d r a t i c e q u a t i o n .$ $i f I h a v e a n e q n . l i k e t h i s$ ${(x^{2} - 1)}^{2} + 2 x (x^{2} - 1) + 1 = 0$ $y o u c a n n o t s o l v e i t l i k e t h i s :$ $x^{2} - 1 = \frac{- 2 x \pm \sqrt{4 x^{2} - 4}}{2} = - x \pm \sqrt{x^{2} - 1}$
	Commented by Tawa1 last updated on 15/Aug/18

	$God bless you sir .$
	Commented by alex041103 last updated on 18/Aug/18

	$T h a t i s n o t a p r o b l e m . I f y o u h a v e s e e n$ $t h e d e r i v a t i o n o f t h e q u a d r a t i c e q u a t i o n,$ $i t i s m a d e b y d o i n g v a l i d a l g e b r a i c$ $m a n i p u l a t i o n . I n o u r c a s e, w e a r e$ $i n f a c t j u s t t r a n s f o r m i n g t h e e q u a t i o n$ $i n t o s o m e o t h e r e q u a t i o n, w e {a r e n}^{'} t s o l v i n g i t .$ $W e j u s t u s e t h e q u a d r a t i c e q u a t i o n$ $t o t r a n s f o r m i t i n t o s o m e o t h e r e q u a t i o n$ $w h i c h i s m o r e f a m i l i a r . . .$
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