Big problem, small solution

Algebra Level 4

log 2 ( 1 + x ) = log 3 x \log _{ 2}{ (1+\sqrt{x})}=\log_{3}{x}

( 1 + a 2 2 a ) y ( 1 a 2 2 a ) y = 1 , { (\frac { 1+a^{ 2 } }{ 2a } ) }^{ y }-{( \frac { 1-a^{ 2 } }{ 2a } )}^{ y }=1, 0 < a 1 0<a\le1

x y = ? \frac { x }{ y } =?


The answer is 4.5.

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3 solutions

Karim Mohamed
Aug 17, 2014

T h e f i r s t p a r t : log 2 ( 1 + x ) = log 3 x = ϕ s o , 3 ϕ = x a n d , 2 ϕ = 1 + x x = ( 2 ϕ 1 ) s o , 2 ϕ 3 ϕ 2 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 : ( 1 + a 2 2 a ) y ( 1 a 2 2 a ) 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 = 1 a , 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 x y = 9 2 = 4.5 The\quad first\quad part:\\ \log _{ 2 }{ (1+\sqrt { x } ) } =\log _{ 3 }{ x } =\phi \\ so,\quad { 3 }^{ \phi }=x\\ and,\quad { 2 }^{ \phi }=1+\sqrt { x } \\ \quad \quad \quad \quad \quad \sqrt { x } ={ ({ 2 }^{ \phi }-1) }\\ so,\quad { 2 }^{ \phi }-{ 3 }^{ \frac { \phi }{ 2 } }-1=0\\ solving\quad for\quad \phi :\quad \phi =2\\ so,\quad x=9\\ The\quad second\quad part\quad is\quad a\quad little\quad longer:\\ { (\frac { 1+{ a }^{ 2 } }{ 2a } ) }^{ y }-{ (\frac { 1-{ a }^{ 2 } }{ 2a } ) }^{ y }=1,\quad 0<a<1\\ we\quad mulitply\quad both\quad sides\quad by\quad { (2a) }^{ y }\\ { (1+{ a }^{ 2 }) }^{ y }-{ (1-{ a }^{ 2 }) }^{ y }={ (2a) }^{ y }\\ we\quad can\quad expand\quad the\quad brackets\quad using\quad binomial\quad theorem:\\ (1+{ C }_{ 1 }^{ y }\times { a }^{ 2 }+{ C }_{ 2 }^{ y }\times { ({ a }^{ 2 }) }^{ 2 }+{ C }_{ 3 }^{ y }\times { ({ a }^{ 2 }) }^{ 3 }+......+{ C }_{ y }^{ y }\times { ({ a }^{ 2 }) }^{ y })-\\ \quad (1-{ C }_{ 1 }^{ y }\times { a }^{ 2 }+{ C }_{ 2 }^{ y }\times { ({ a }^{ 2 }) }^{ 2 }+{ -C }_{ 3 }^{ y }\times { ({ a }^{ 2 }) }^{ 3 }+......+{ (-1) }^{ y }{ C }_{ y }^{ y }\times { ({ a }^{ 2 }) }^{ y })\\ ={ (2a) }^{ y }\quad .......(1)\\ the\quad odd\quad orderd\quad terms\quad in\quad the\quad two\quad brackets\quad will\quad \\ cancel\quad each\quad other.\quad the\quad problem\quad is\quad we\quad don't\\ know\quad (y)\quad is\quad even\quad or\quad odd.\\ \\ (i)\quad let\quad (y)\quad be\quad odd:\quad this\quad makes\quad the\quad last\quad term\quad in\\ the\quad second\quad bracket\quad a\quad negative\quad term\quad so\quad it\quad will\\ not\quad be\quad cancelled.\\ so\quad the\quad last\quad equation\quad gives:\\ 2{ a }^{ 2 }({ C }_{ 1 }^{ y }+{ C }_{ 3 }^{ y }\times { ({ a }^{ 2 }) }^{ 2 }+{ C }_{ 5 }^{ y }\times { ({ a }^{ 2 }) }^{ 4 }+....+{ C }_{ y }^{ y }\times { ({ a }^{ 2 }) }^{ y })={ (2a) }^{ y }\\ { C }_{ 1 }^{ y }+{ C }_{ 3 }^{ y }\times { ({ a }^{ 2 }) }^{ 2 }+{ C }_{ 5 }^{ y }\times { ({ a }^{ 2 }) }^{ 4 }+....+{ C }_{ y }^{ y }\times { ({ a }^{ 2 }) }^{ y }\quad =\quad { 2 }^{ y-1 }\quad \times \quad { a }^{ y-2 }\\ By\quad trial\quad and\quad error:\\ -\quad y=1:\quad \quad 1=\frac { 1 }{ a } ,\quad so\quad a=1\quad (refused)\\ -\quad y=3:\quad \quad 3+{ a }^{ 4 }=4a,\quad so\quad the\quad real\quad root\quad is\quad 1\quad (refused)\\ if\quad we\quad continue\quad the\quad value\quad of\quad (a)\quad will\quad diverge\quad (\left| a \right| >1)\\ so\quad (y)\quad can\quad not\quad be\quad odd.\\ \\ (ii)\quad let\quad (y)\quad be\quad even:\\ \quad the\quad last\quad term\quad in\quad the\quad second\quad bracket\quad will\quad be\\ \quad cancelled\quad and\quad equation\quad (1)\quad gives:\\ { C }_{ 1 }^{ y }+{ C }_{ 3 }^{ y }\times { ({ a }^{ 2 }) }^{ 2 }+{ C }_{ 5 }^{ y }\times { ({ a }^{ 2 }) }^{ 4 }+.....+{ C }_{ y-1 }^{ y }\times { ({ a }^{ 2 }) }^{ y-1 }\quad =\quad { 2 }^{ y-1 }\times { a }^{ y-2 }\\ By\quad trial\quad and\quad error:\\ -\quad y=2:\quad 2={ 2 }^{ 1 }\times { a }^{ 0 }\quad \quad valid\\ -\quad y=4:\quad 4+4{ a }^{ 4 }=8{ a }^{ 2 },\quad the\quad real\quad roots\quad are\quad 1,-1\quad (refused)\\ if\quad we\quad continue\quad the\quad value\quad of\quad (a)\quad will\quad diverge\quad (\left| a \right| >1)\\ so\quad y=2\\ \\ \frac { x }{ y } =\frac { 9 }{ 2 } =4.5

Excellent solution! As good as original!

John M. - 6 years, 9 months ago

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thanks john

Karim Mohamed - 6 years, 9 months ago

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Um, could you please explain how you solved 2 x 3 x / 2 1 = 0 ? 2^x-3^{x/2}-1=0?

Zero Mech - 6 years, 9 months ago

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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

Karim Mohamed - 6 years, 9 months ago

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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 \phi 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 2^x-3^{x/2}-1=0 solution too.

But still, gj on the second one.

John M. - 6 years, 9 months ago

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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 ϕ 2 1 = 0 { 2 }^{ \phi }-{ 3 }^{ \frac { \phi }{ 2 } }-1=0 , and got the answer ϕ = 2 \phi =2 .

What's the other way to solve such type of exponential equations ?

Karim Mohamed - 6 years, 9 months ago

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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.

John M. - 6 years, 9 months ago
Nguyen Thanh Long
Aug 13, 2014

Assume: log 2 ( 1 + x ) = log 3 x = t \log_2(1+\sqrt{x})=\log_3{x}=t 1 + 3 t = 2 t t = 2 , x = 9 \Rightarrow 1+\sqrt{3^t}=2^t \Rightarrow t=2, x=9 It is easy to see y=2 satisfies the second equation. x y = 4.5 \frac{x}{y}=\boxed{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.

John M. - 6 years, 10 months ago

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Not exactly a good job on the first one either.. they didn't explain how they got t = 2 t = 2 . But then again, it's hard to solve the first one without guessing the answer.

Ariel Gershon - 6 years, 10 months ago
Hasan Kassim
Aug 24, 2014

As others did, I solved the first equation by guessing...

The second one can be solved as follows:

let a = tan x \displaystyle a=\tan x for x ] 0 , π 4 [ \displaystyle x\in ]0,\frac{\pi}{4}[

Then using trigonometry:

( 1 + a 2 2 a ) y ( 1 a 2 2 a ) y = 1 \displaystyle (\frac{1+a^2}{2a})^y - (\frac{1-a^2}{2a})^y=1

< = > ( 1 + tan 2 x 2 tan x ) y ( 1 tan 2 x 2 tan x ) y = 1 \displaystyle <=> (\frac{1+\tan^2x}{2\tan x})^y- (\frac{1-\tan^2x}{2\tan x})^y=1

< = > ( 1 sin 2 x ) y ( cos 2 x sin 2 x ) y = 1 \displaystyle <=> (\frac{1}{\sin 2x})^y- (\frac{\cos 2x}{\sin 2x})^y=1

< = > 1 cos y 2 x = sin y 2 x \displaystyle <=> 1-\cos^y2x=\sin^y2x

The last equation is true only for y = 2 y=2 which corresponds to the pythagorian identity.

also, The result shows that the solution is true for all a a not for just 0 < a < 1 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.

John M. - 6 years, 9 months ago

Ow, just noticed the last part of your solution; it isn't right.

The solution is not true for all a a . When you made the substitution, you made a promise to keep tan x \tan {x} between 0 0 and π 4 \frac{\pi}{4} .

Also, on your transition from tan on denominator to sin 2x on denominator, you cancel cos x \cos{x} , and there you promise that cos x 0 \cos{x} \neq 0 . Since sin x a = c o s x \frac{\sin{x}}{a}=cos{x} , this implies that s i n x 0 sin{x} \neq 0 (which agrees with the original condition that tan x \tan{x} is bound between 0 and π 4 \frac{\pi}{4} ). Also, when you multiply through by sin y 2 x \sin^y{2x} , you also promise it to not equal 0.

Without furthermore complex reasoning, we can see that a a cannot exceed 1. If a = 2 a=2 , then we obtain ( 5 4 ) 2 ( 1 4 ) 2 = 24 16 1 (\frac{5}{4})^2-(\frac{1}{4})^2=\frac{24}{16} \neq 1 . In terms of your solution, the last expression may be rewritten as

sin y 2 x + cos y 2 x = 1 \sin^y{2x}+\cos^y{2x}=1

( 2 tan x ) y + [ ( 1 t a n 2 x ) ( cos 2 x ) ] y = 1 (2\tan{x})^y+[(1-tan^2{x})(\cos^2{x})]^y=1

( 2 a ) y + ( 1 a 2 ) y ( cos 2 y x ) = 1 (2a)^y+(1-a^2)^y(\cos^{2y}{x})=1

Now, if a > 1 a>1 , it is obvious that since y = 2 y=2 , ( 2 a ) y + ( 1 a 2 ) y ( cos 2 y x ) > 2 1 (2a)^y+(1-a^2)^y(\cos^{2y}{x})>2 \neq 1 . for all x x and for all a a greater than 1 or equal to 1.

Similar reasoning will reveal that a < 0 a<0 is a no-no.

However, plugging in a = 1 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 0 \le a \le 1 . But then again, a 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

John M. - 6 years, 9 months ago

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Well John I think you committed some calculation errors while disproving my conclusion...

First of all replace y = 2 y=2 in the second equation, you will simplify to one, discarding the zero value of a 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 \sin 2x .

Also, when you substituted a = 2 a=2 , you made a calculation mistake.

Anyway thanks for pointing out that a a should be non zero :)

Hasan Kassim - 6 years, 9 months ago

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