Algebra Problem of The Day-3

[BigBeachBanana]

[math]9^x+15^x=25^x[/math]Solve for all [imath]x \in \R[/imath].

Extra fun: Solve for all [imath]x \in \mathbb{C}[/imath]

 

[Steven G]

Spoiler

9^x + 15^x + 25^x
1 + (5/3)^x = [(5/3)^x]^2
The rest is clear
x=ln((1+sqrt(5))/2)/ln(5/3)

 

[Cubist]

I'm still having a think about the complex number case! Here's my method for the real x, which is basically the same as @Steven G #2


And I spy the golden ratio! Here's the method in Cubist (baby ) steps, with exactly the same answer as #2 but given in terms of phi...

9^x + 15^x = 25^x
9^x + (5*3)^x - (5*5)^x = 0

Divide by 9^x

1 + (5*3/(3*3))^x - (5*5/(3*3))^x = 0
1 + (5/3)^x - ((5/3)*(5/3))^x = 0

Let u = (5/3)^x AA

u^2 - u - 1 = 0

Quadratic formula

u = (1 ± sqrt(5))/2

For real x, u = (1 + sqrt(5))/2 = phi, [imath]\phi[/imath], the golden ratio. Using AA to obtain x...

[math]x = \log_{(5/3)}\phi \approx 0.94202757784924[/math]

 

[Cubist]

Extra fun: Solve for all [imath]x \in \mathbb{C}[/imath]

Spoiler: method

Continuing from the quadratic formula of post#3
  u = (1 + sqrt(5))/2 = phi       B1
  u = (1 - sqrt(5))/2 = -1/phi    B2

Let x=r*e^(i*theta)   thanks Euler!
Let c=5/3

Since u = (5/3)^x then
u = c^x
  = c^(r*e^(i*theta))
  = c^(r*cos(theta) + r*i*sin(theta))
  = c^(r*cos(theta)) * c^(i*r*sin(theta))
  = c^(r*cos(theta)) * e^( (i*r*sin(theta)) * ln(c) ) since c>0
  = c^(r*cos(theta)) * (  cos(r*sin(theta)*ln(c)) + i*sin(r*sin(theta)*ln(c)) )
  equation ZZ

And since B1 & B2 dictate that the imaginary part of u must be 0 AND the real part is NOT 0 then ZZ implies...
  sin(r*sin(theta)*ln(c)) = 0
  r = I*pi/(ln(c)*sin(theta))    CC

Continue from ZZ but now without the imaginary part...
u = c^(r*cos(theta))*cos(r*sin(theta)*ln(c))

Using CC's value of r...

u = c^((I*pi/(ln(c)*sin(theta)))*cos(theta)) * cos((I*pi/(ln(c)*sin(theta)))*ln(c)*sin(theta))
Note   cos(  I*pi * (ln(c)*sin(theta))/(ln(c)*sin(theta)) )
     = cos(  I*pi )
     = (-1)^I
u = c^(I*pi*cos(theta)/(ln(c)*sin(theta))) * (-1)^I
And since tan = sin/cos
u = c^(I*pi/(tan(theta)*ln(c))) * (-1)^I

B1 and B2 imply that u has a specific, real, value. Let this value be x (which will be one of the two values x=phi or x=-1/phi)

c^(I*pi/(tan(theta)*ln(c)))*(-1)^I = x
I*pi*ln(c)/(tan(theta)*ln(c)) = ln((-1)^I*x)
tan(theta) = I*pi/ln((-1)^I*x)
theta = atan(I*pi/ln((-1)^I*x))   DD

--

NOW HOW DO WE USE ALL OF THAT ???

Choose a non zero value of I (otherwise equation CC fails, not sure if this merits further research!)
For DD to be valid...
  if I is even then x=phi
  if I is odd  then x=-1/phi
DD implies  theta = atan(I*pi/ln((-1)^I*x))
CC implies  r = I*pi/(ln(5/3)*sin(theta))
FINALLY x = r*e^(i*theta)

Spoiler: some of the infinite complex values for x

I've checked some of these x values in Octave (which is similar to Matlab) and they seem ok

 I    Approximate x
===========================================================
-1    -6.22175849434065812051*e^(i* 1.41880320712571009594)
 1    -6.22175849434065812051*e^(i*-1.41880320712571009594)
 
-2    12.33607989497940200292*e^(i*-1.49435830486998915870)
 2    12.33607989497940200292*e^(i* 1.49435830486998915870)
 
-3   -18.47412193306283091922*e^(i* 1.51978246555061253334)
 3   -18.47412193306283091922*e^(i*-1.51978246555061253334)
 
-4    24.61814820064609443585*e^(i*-1.53252140796599722864)
 4    24.61814820064609443585*e^(i* 1.53252140796599722864)

-5   -30.76457355578359352938*e^(i* 1.54017100873374241212)
 5   -30.76457355578359352938*e^(i*-1.54017100873374241212)

 

[Cubist]

After plotting some of the complex solutions that I found above, I felt sure that a simpler formula would be possible...

Spoiler: simple formula

By extending the work in post#4 then clearly(*)...
x = ( (-1)^I*ln(phi) + i*I*pi )/ln(5/3)
where I is an integer (negative values are allowed)

* - It took me quite a few lines of messy looking work to get the result But, given how simple the final answer is there must surely be a short way to find this from the original equation?

 

[BigBeachBanana]

After plotting some of the complex solutions that I found above, I felt sure that a simpler formula would be possible...

Spoiler: simple formula

By extending the work in post#4 then clearly(*)...
x = ( (-1)^I*ln(phi) + i*I*pi )/ln(5/3)
where I is an integer (negative values are allowed)

* - It took me quite a few lines of messy looking work to get the result But, given how simple the final answer is there must surely be a short way to find this from the original equation?

Well done!