Factorization Is An Art

Algebra Level 2

201 6 x + 201 6 x = 3 \large 2016^{x}+2016^{-x}=3

201 6 6 x 201 6 6 x 201 6 x 201 6 x = ? \large \sqrt{\frac{2016^{6x}-2016^{-6x}}{2016^{x}-2016^{-x}}} = \, ?


The answer is 12.

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3 solutions

Chew-Seong Cheong
Aug 22, 2016

Let a = 201 6 x a=2016^x . Then 201 6 x + 201 6 x = a + a 1 = 3 2016^x + 2016^{-x} = a+a^{-1} = 3 and we have:

X = 201 6 6 x 201 6 6 x 201 6 x 201 6 x = a 6 a 6 a a 1 = ( a 3 a 3 ) ( a 3 + a 3 ) a a 1 = ( a a 1 ) ( a 2 + a 2 + 1 ) ( a + a 1 ) ( a 2 + a 2 1 ) a a 1 = ( a 2 + a 2 + 1 ) ( 3 ) ( a 2 + a 2 1 ) = 3 ( ( a + a 1 ) 2 1 ) ( ( a + a 1 ) 2 3 ) = 3 ( 9 1 ) ( 9 3 ) = 3 ( 8 ) ( 6 ) = 144 = 12 \begin{aligned} X & = \sqrt {\frac{2016^{6x}-2016^{-6x}}{2016^x - 2016^{-x}}} \\ & = \sqrt{\frac {a^6-a^{-6}}{a-a^{-1}}} \\ & = \sqrt{\frac {\color{#3D99F6}{(a^3-a^{-3})} \color{#D61F06}{(a^3 + a^{-3})}}{a-a^{-1}}} \\ & = \sqrt{\frac {\color{#3D99F6}{\cancel{(a-a^{-1})}(a^2+a^{-2}+1)} \color{#D61F06}{(a + a^{-1})(a^2+a^{-2}-1)}}{\cancel{a-a^{-1}}}} \\ & = \sqrt{\color{#3D99F6}{(a^2+a^{-2}+1)}\color{#D61F06}{(3)(a^2+a^{-2}-1)}} \\ & = \sqrt{{\color{#D61F06}3} {\color{#3D99F6}\left(\left(a+a^{-1}\right)^2 -1 \right)} {\color{#D61F06}\left(\left(a+a^{-1}\right)^2 -3 \right)}} \\ & = \sqrt{{\color{#D61F06}3} {\color{#3D99F6}\left(9 -1 \right)} {\color{#D61F06}\left(9 -3 \right)}} = \sqrt{{\color{#D61F06}3} {\color{#3D99F6}\left(8 \right)} {\color{#D61F06}\left(6 \right)}} = \sqrt{144} = \boxed{12} \end{aligned}

What a hard problem! I thought too much about it!

Tấn Phát Nguyễn - 4 years, 9 months ago

From the 5th line, I think it would be clearer to factorise the a 2 + a 2 + 1 a^2+a^{-2}+1 and a 2 + a 2 1 a^2+a^{-2}-1 in terms of perfect squares. Aside from that, nice answer!

A Former Brilliant Member - 4 years, 7 months ago

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Thanks. I have changed the solution.

Chew-Seong Cheong - 4 years, 7 months ago
Zach Bian
Sep 3, 2016

Python numerical root finder:

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from math import *
from scipy.misc import derivative

# Newton's method for finding roots
# Original function
f = lambda x: (2016**x)+(2016**-x)-3

roots=[]
guesses=[]

bound = 1     # 2016 >> 3. Loose upper bound.
i=-bound         
while i <= bound:
    guesses.append(i)
    i+=0.5

for guess in guesses:
    val = f(guess)
    # deriv = fp(guess)
    deriv = derivative(f, guess, 1e-3)
    if deriv == 0:
        continue
    nextguess = -val/deriv
    err = 1e-5

    while fabs(nextguess-guess)>err and fabs(nextguess-guess)<1e12:
        guess = nextguess
        deriv = derivative(f, guess, 1e-3)
        if deriv == 0:
            continue
        nextguess = guess-f(guess)/deriv

    rounded = round(nextguess, 10)
    if rounded not in roots:
        roots.append(rounded)

acc = 0
for root in roots:
    # Dummy check:
    if fabs(f(root)) < 1e-9:
        print(root)
        acc += root

x=roots[0]
print(sqrt((2016**(6*x)-2016**(-6*x))/(2016**x-2016**-x)))

Ramiel To-ong
Sep 1, 2016

nice solution

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