Slightly-modified-form-of-Q-1343-3-log-3x-4-4-log-4x-3-solve-for-x-

Question Number 1354 by Rasheed Ahmad last updated on 25/Jul/15

S l i g h t l y m o d i f i e d f o r m o f Q 1343 .

3^{l o g (3 x + 4)} = 4^{l o g (4 x + 3)}, s o l v e f o r x .

Commented by prakash jain last updated on 25/Jul/15

One option to prove that x = 0 is the only

solution for \frac{\log (3 x + 4)}{\log (4 x + 3)} - \frac{\log 4}{\log 3} = 0 is to

take derivative of the function .

Also if the function has more than one zero

it must change sign twice .

Answered by Yugi last updated on 25/Jul/15

3^{l o g (3 x + 4)} = 4^{l o g (4 x + 3)} \dots \dots \dots (1)

T a k i n g l o g s t o b a s e 10 o n b o t h s i d e s o f (1)

[l o g (3 x + 4)] l o g 3 = [l o g (4 x + 3)] l o g 4

\Rightarrow \frac{l o g (3 x + 4)}{l o g (4 x + 3)} = \frac{l o g 4}{l o g 3} \dots \dots \dots . . (2)

U p o n o b s e r v a t i o n o f t h e f o r m o f (2), t h e o n l y r e a l

s o l u t i o n i s x = 0 . A g r a p h i c a l a p p r o a c h y i e l d s

t h e s a m e r e s u l t s i n c e (1) i s e q u i v a l e n t t o y = 0 f o r

y = 3^{l o g (3 x + 4)} - 4^{l o g (4 x + 3)} d e f i n e d f o r x > - \frac{3}{4} .

N u m e r i c a l m e t h o d s w o r k a s w e l l .

O t h e r w i s e, I c o u l d n o t f i n d a n a l g e b r a i c m e a n s

o f s o l v i n g f o r x o f t h e f o r m

\frac{l o g (a x + b)}{l o g (c x + d)} = e w h e r e a l l o f a, b, c, d a n d e a r e

n o n - z e r o r e a l s .

Commented by Rasheed Soomro last updated on 10/Aug/15

I a p p r e c i a t e y o u r a p p r o a c h . B u t I a m d o u b t f u l a b o u t y o u r

c l a i m t h a t t h e r e i s o n l y o n e r e a l s o l u t i o n (x = 0) .

Y o u c o r r e c t l y d e r i v e d :

\frac{l o g (3 x + 4)}{l o g (4 x + 3)} = \frac{l o g 4}{l o g 3}

T a k i n g f e w s t e p s m o r e, w e w i l l g e t,

l o g (3 x + 4) = \frac{l o g 4}{l o g 3} . l o g (4 x + 3)

= l o g {(4 x + 3)}^{\frac{l o g 4}{l o g 3}}

T a k i n g a n t i l o g o f b o t h s i d e s

3 x + 4 = {(4 x + 3)}^{\frac{l o g 4}{l o g 3}}

O b s e r v e r i g h t h a n d s i d e t h e i n d e x \frac{l o g 4}{l o g 3} > 0 .

S o f o r p o s i t i v e 4 x + 3 t h e R H S m u s t b e r e a l, i f I a m

n o t w r o n g . S o I t h i n k t h e r e m a y b e o t h e r r e a l s o l u t i o n / s t o o .

I a g r e e w i t h y o u t h a t t h e r e i s n o a l g e b r a i c w a y t o s o l v e

t h i s e q u a t i o n . B u t n u m e r i c a l m e t h o d s m a y h e l p .

Commented by Yugi last updated on 25/Jul/15

C o n s i d e r i n g t h e f o r m

3 x + 4 = {(4 x + 3)}^{\frac{l o g 4}{l o g 3}} \dots \dots . . (*)

\frac{l o g 4}{l o g 3} = {l o g}_{3} 4 i s a n i r r a t i o n a l e x p o n e n t . A s s u c h, t h e f o l l o w i n g n u m e r i c a l a p p r o a c h t o

s o l v i n g f o r x c a n b e d o n e i n o r d e r t o a c q u i r e a p o l y n o m i a l i n x (w h o s e r e a l s o l u t i o n

s h o u l d b e f o u n d t o b e a p p r o x i m a t e l y x = 0) . F i r s t l y, c o n s i d e r t h e v a l u e o f {l o g}_{3} 4 .

{l o g}_{3} 4 = 1.2618 \dots S o t o 2 s . f {l o g}_{3} 4 \approx 1.3 = \frac{13}{10}

∴ 3 x + 4 = {(4 x + 3)}^{13 / 10}

R a i s i n g b o t h s i d e s b y t h e p o w e r o f 10 g i v e s

{(3 x + 4)}^{10} = {(4 x + 3)}^{13}

W h i l e i t i s p o s s i b l e t o e x p a n d t h o s e b i n o m i a l s t o s o l v e f o r x i n f (x) = 0 I t h i n k i t w o u l d b e e a s i e r t o d e f i n e

a f u n c t i o n y = {(3 x + 4)}^{10} - {(4 x + 3)}^{13} (x > - 3 / 4, x \in R) a n d u s e n u m e r i c a l m e t h o d s

f o r e s t i m a t i n g i t s r o o t . G r a p h i n g y i e l d s a n a n s w e r t h e f a s t e s t .

N o w, t h i s g i v e s o n e e s t i m a t i o n f o r x s a t i s f y i n g (*) . T o a t t a i n a m o r e a c c u r a t e

e s t i m a t e, u s e m o r e a c c u a r t e r a t i o n a l a p p r o x i m a t i o n s t o {l o g}_{3} 4, s u c h a s \frac{63}{50}, \frac{6309}{5000}, e t c .

E v e r y r e s u l t o f x b a s e d o n t h e s e a p p r o x i m a t i o n s h a v e d i f f e r e n t v a l u e s

b u t t h e a n s w e r s a r e n e a r z e r o . W h a t I^{'} v e o u t l i n e d h e r e i s o n e m e a n s

o f o b t a i n i n g x w i t h o u t k n o w i n g h o w m a n y r e a l r o o t s e x i s t f o r (*) . H o w e v e r,

i n g e n e r a l, i f x i s a n e g a t i v e r e a l n u m b e r a n d y i s i r r a t i o n a l, t h e n a l l a n s w e r s o f

x^{y} a r e c o m p l e x . A l s o, i f x i s p o s i t i v e (x \in R) a n d y \in \bar{Q} t h e n x^{y} h a s o n l y o n e r e a l

a n s w e r a n d a l l o f t h e r e s t a r e c o m p l e x . H e n c e, R H S (*) s h o u l d y i e l d o n e r e a l r e s u l t .

(O f c o u r s e I c o u l d b e w r o n g . I d i d s o m e r e s e a r c h b u t I^{'} m j u s t a s t u d e n t {w h o}^{'} s h e r e

t o l e a r n .)

Commented by Rasheed Soomro last updated on 26/Jul/15

I {d i d n}^{'} t s a y^{″} T h e r e a r e certainly m o r e t h a n o n e s o u t i o n^{″} .

B u t I s a i d,^{″} T h e r e may be m o r e t h a n o n e {s o l u t i o n}^{″} a n d

s u g g e s t e d t o c o n f i r m u s i n g n u m e r i c a l a n a l y s i s . O n e o t h e r

s u g g e s t i o n h a s c o m e f r o m p a r k a s h j a i n f o r t h e c o n f i r m a t i o n .

B u t d u r i n g t h i s c o m m e n t I h a v e d r a w n t h e g r a p h . It seems that your claim of^{″} only

one real solution is correct^{″} .

A f t e r a l l w e a l l a r e l e a r n e r s .

Commented by Yugi last updated on 26/Jul/15

Y e s . T h e g r a p h s h o w e d t h e e x i s t e n c e o f o n e r e a l s o l u t i o n .

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