What is $x$, if $3^x+3^{-x}=1$?

Hints (why it's impossible in reals):

  • $3^{x} \gt 0$ for any real $\forall x \in \mathbb{R}$

  • $a + \cfrac{1}{a} \ge 2$ for any positive real $\forall a \in \mathbb{R}^+$


Note that we have $$ 3^x+3^{-x}=(3^{x/2}-3^{-x/2})^2+2 $$ Apply this to your equation, and you get $$ (3^{x/2}-3^{-x/2})^2=-1 $$ Which means that $3^{x/2}-3^{-x/2}$ is imaginary.


You have proceeded wrongly. We have $$3^x +3^{-x} =1 $$ $$\Rightarrow 3^{2x} +1 =3^x $$ $$\Rightarrow 3^{2x} -3^x +1=0$$ giving us $$3^x = \frac {1\pm \sqrt {3}i}{2}$$

Notice that the LHS is always real but we have an imaginary part on the RHS. This is enough evidence to suggest that the above equation has no solutions. Hope it helps.

Tags:

Algebra Precalculus

Quadratics

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