Matematické Fórum

stuart clark · 22. 01. 2015 09:34

No. of real solution of the equation $1+8^x+27^x = 2^x+12^x+9^x$

Eratosthenes · 22. 01. 2015 19:33

↑ stuart clark:

Two.

Marian · 30. 01. 2015 13:31 — Editoval Marian (30. 01. 2015 13:34)

↑ stuart clark:

There is only one real solution. I found the following algebraic approach....

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stuart clark · 08. 02. 2015 13:56

Thanks ↑ Marian:

Pavel · 15. 02. 2015 14:34 — Editoval Pavel (15. 02. 2015 14:36)

↑ stuart clark:

I have another solution that uses a special property concerning of six positive real numbers, see here. (The problem was solved sooner.)

If $a,b,c,x,y,z\in\mathbb R^+$ such that

1. $a\leq x < y < z \leq c$ ,
2. $a < b < c$ ,
3. $a+b+c = x+y+z$ ,
4. $abc = xyz$

then $a=x$ , $b=y$ , $c=z$ .

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It is clear that $s=0$ is a solution of the equation $1+8^s+27^s=2^s+12^s+9^s$ . We show that there is the only solution.

Let $s>0$ and $a=1$ , $b=8^s$ , $c=27^s$ , $x=2^s$ , $y=9^s$ and $z=12^s$ . All the four properties above are satisfied.

Then $1=2^s$ , $8^s=9^s$ and $27^s=12^s$ . However, the equations have no solution.

We can deduce the same conlusion, if $s<0$ and $a=27^s$ , $b=8^s$ , $c=1$ , $x=12^s$ , $y=9^s$ and $z=2^s$ .