lo g 2 ( 1 + x ) = lo g 3 x
( 2 a 1 + a 2 ) y − ( 2 a 1 − a 2 ) y = 1 , 0 < a ≤ 1
y x = ?
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Excellent solution! As good as original!
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thanks john
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Um, could you please explain how you solved 2 x − 3 x / 2 − 1 = 0 ?
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@Zero Mech – i solved it online on wolfram, you can- if you want- solve it by newton's method, or graphically on any mathematics software. if you have a way to solve it analytically on hand, please share it
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@Karim Mohamed – Oh wait I didn't look at the entire solution;
Yes, you need to show how you solved for ϕ in the first part. Otherwise the solution is incomplete.
Just to point out, there is another way to solve this. But I'd like to see the 2 x − 3 x / 2 − 1 = 0 solution too.
But still, gj on the second one.
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@John M. – The free wolfram software does not provide a step by step solution. I wrote the equation 2 ϕ − 3 2 ϕ − 1 = 0 , and got the answer ϕ = 2 .
What's the other way to solve such type of exponential equations ?
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@Karim Mohamed – Oh well this I dunno :O played around with it for a while and - nothing.
I'll post it as a note, let's see if anyone can do it.
Assume: lo g 2 ( 1 + x ) = lo g 3 x = t ⇒ 1 + 3 t = 2 t ⇒ t = 2 , x = 9 It is easy to see y=2 satisfies the second equation. y x = 4 . 5
It may be easy to see that y=2 satisfies the second equation, but the point is to solve it rather than apply a brute force method.
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Not exactly a good job on the first one either.. they didn't explain how they got t = 2 . But then again, it's hard to solve the first one without guessing the answer.
As others did, I solved the first equation by guessing...
The second one can be solved as follows:
let a = tan x for x ∈ ] 0 , 4 π [
Then using trigonometry:
( 2 a 1 + a 2 ) y − ( 2 a 1 − a 2 ) y = 1
< = > ( 2 tan x 1 + tan 2 x ) y − ( 2 tan x 1 − tan 2 x ) y = 1
< = > ( sin 2 x 1 ) y − ( sin 2 x cos 2 x ) y = 1
< = > 1 − cos y 2 x = sin y 2 x
The last equation is true only for y = 2 which corresponds to the pythagorian identity.
also, The result shows that the solution is true for all a not for just 0 < a < 1 .
Woah you really have a way with trig, don't you?
But I guess I'll just have to post the solution for the first one. (As soon as I get time to come up with one in accords to my book, that is).
Good job.
Ow, just noticed the last part of your solution; it isn't right.
The solution is not true for all a . When you made the substitution, you made a promise to keep tan x between 0 and 4 π .
Also, on your transition from tan on denominator to sin 2x on denominator, you cancel cos x , and there you promise that cos x = 0 . Since a sin x = c o s x , this implies that s i n x = 0 (which agrees with the original condition that tan x is bound between 0 and 4 π ). Also, when you multiply through by sin y 2 x , you also promise it to not equal 0.
Without furthermore complex reasoning, we can see that a cannot exceed 1. If a = 2 , then we obtain ( 4 5 ) 2 − ( 4 1 ) 2 = 1 6 2 4 = 1 . In terms of your solution, the last expression may be rewritten as
sin y 2 x + cos y 2 x = 1
( 2 tan x ) y + [ ( 1 − t a n 2 x ) ( cos 2 x ) ] y = 1
( 2 a ) y + ( 1 − a 2 ) y ( cos 2 y x ) = 1
Now, if a > 1 , it is obvious that since y = 2 , ( 2 a ) y + ( 1 − a 2 ) y ( cos 2 y x ) > 2 = 1 . for all x and for all a greater than 1 or equal to 1.
Similar reasoning will reveal that a < 0 is a no-no.
However, plugging in a = 1 directly into the original reveals a true value. I don't know what to make of THAT. I think I misinterpreted the symbols; it should be 0 ≤ a ≤ 1 . But then again, a can't be zero. So I'll edit the problem to include the 1.
I may have misanalyzed something, tell me if I did.
Cheers
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Well John I think you committed some calculation errors while disproving my conclusion...
First of all replace y = 2 in the second equation, you will simplify to one, discarding the zero value of a .
Secondly, when you proved that the equation is greater than 2 for all a and x, you made a mistake concerning the substitution for sin 2 x .
Also, when you substituted a = 2 , you made a calculation mistake.
Anyway thanks for pointing out that a should be non zero :)
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T h e f i r s t p a r t : lo g 2 ( 1 + x ) = lo g 3 x = ϕ s o , 3 ϕ = x a n d , 2 ϕ = 1 + x x = ( 2 ϕ − 1 ) s o , 2 ϕ − 3 2 ϕ − 1 = 0 s o l v i n g f o r ϕ : ϕ = 2 s o , x = 9 T h e s e c o n d p a r t i s a l i t t l e l o n g e r : ( 2 a 1 + a 2 ) y − ( 2 a 1 − a 2 ) y = 1 , 0 < a < 1 w e m u l i t p l y b o t h s i d e s b y ( 2 a ) y ( 1 + a 2 ) y − ( 1 − a 2 ) y = ( 2 a ) y w e c a n e x p a n d t h e b r a c k e t s u s i n g b i n o m i a l t h e o r e m : ( 1 + C 1 y × a 2 + C 2 y × ( a 2 ) 2 + C 3 y × ( a 2 ) 3 + . . . . . . + C y y × ( a 2 ) y ) − ( 1 − C 1 y × a 2 + C 2 y × ( a 2 ) 2 + − C 3 y × ( a 2 ) 3 + . . . . . . + ( − 1 ) y C y y × ( a 2 ) y ) = ( 2 a ) y . . . . . . . ( 1 ) t h e o d d o r d e r d t e r m s i n t h e t w o b r a c k e t s w i l l c a n c e l e a c h o t h e r . t h e p r o b l e m i s w e d o n ′ t k n o w ( y ) i s e v e n o r o d d . ( i ) l e t ( y ) b e o d d : t h i s m a k e s t h e l a s t t e r m i n t h e s e c o n d b r a c k e t a n e g a t i v e t e r m s o i t w i l l n o t b e c a n c e l l e d . s o t h e l a s t e q u a t i o n g i v e s : 2 a 2 ( C 1 y + C 3 y × ( a 2 ) 2 + C 5 y × ( a 2 ) 4 + . . . . + C y y × ( a 2 ) y ) = ( 2 a ) y C 1 y + C 3 y × ( a 2 ) 2 + C 5 y × ( a 2 ) 4 + . . . . + C y y × ( a 2 ) y = 2 y − 1 × a y − 2 B y t r i a l a n d e r r o r : − y = 1 : 1 = a 1 , s o a = 1 ( r e f u s e d ) − y = 3 : 3 + a 4 = 4 a , s o t h e r e a l r o o t i s 1 ( r e f u s e d ) i f w e c o n t i n u e t h e v a l u e o f ( a ) w i l l d i v e r g e ( ∣ a ∣ > 1 ) s o ( y ) c a n n o t b e o d d . ( i i ) l e t ( y ) b e e v e n : t h e l a s t t e r m i n t h e s e c o n d b r a c k e t w i l l b e c a n c e l l e d a n d e q u a t i o n ( 1 ) g i v e s : C 1 y + C 3 y × ( a 2 ) 2 + C 5 y × ( a 2 ) 4 + . . . . . + C y − 1 y × ( a 2 ) y − 1 = 2 y − 1 × a y − 2 B y t r i a l a n d e r r o r : − y = 2 : 2 = 2 1 × a 0 v a l i d − y = 4 : 4 + 4 a 4 = 8 a 2 , t h e r e a l r o o t s a r e 1 , − 1 ( r e f u s e d ) i f w e c o n t i n u e t h e v a l u e o f ( a ) w i l l d i v e r g e ( ∣ a ∣ > 1 ) s o y = 2 y x = 2 9 = 4 . 5