Solutions?

Algebra Level 4

How many solutions for x x exist satisfying 3 2 x + ( 2 × 3 x ) + 1 = 0 3^{2x}+(2\times 3^{x})+1=0 ?

Note: solutions may be real or complex.

Infinitely many 1 2 0

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1 solution

Miles Koumouris
Apr 14, 2017

Let 3 x = a 3^{x}=a ; then

a 2 + 2 a + 1 = 0 ( a + 1 ) 2 = 0 a = 1 a^{2}+2a+1=0\Longrightarrow (a+1)^{2}=0\Longrightarrow a=-1 .

So 3 x = 1 3^{x}=-1\Longrightarrow no real solutions exist.

But now let 3 x = 1 = e i ( 2 k + 1 ) π , k Z 3^{x}=-1=e^{i(2k+1)\pi }, k\in \mathbb{Z} ;

then ln ( 3 x ) = i ( 2 k + 1 ) π x = i ( 2 k + 1 ) π ln ( 3 ) , k Z \textrm{ln}(3^{x})=i(2k+1)\pi \Longrightarrow x=\dfrac{i(2k+1)\pi }{\textrm{ln}(3)}, k\in \mathbb{Z} .

So infinitely many solutions exist.

Damnit, I included only the case where k = 0.

Konstantin Zeis - 4 years, 1 month ago

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