Equations and Inequations

The above exponential equation can be rewritten as $(3^3)^{\frac{1}{x}} + (2^{2}3^{1})^{\frac{1}{x}} = 2(2^3)^{\frac{1}{x}}$ , or $(3^{\frac{1}{x}})^3 + (3^{\frac{1}{x}})(2^{\frac{1}{x}})^2 = 2(2^{\frac{1}{x}})^3.$ Now, let $u = 3^{\frac{1}{x}}, v = 2^{\frac{1}{x}}$ so that we obtain the quadratic:

$u^3 + uv^2 = 2v^3$ ;

or $u^3 - v^3 = v^3 - uv^2$ ;

or $(u-v)(u^2 + uv + v^2) = -v^2 (u-v)$ ;

or $u^2 + uv + 2v^2 = 0$ ;

or $u = \frac{-v \pm \sqrt{v^2 - 4(1)(2v^2)}}{2} = \frac{-v \pm \sqrt{-7v^2}}{2} = (\frac{-1 \pm \sqrt{7}i}{2}) \cdot v$ ;

or $\frac{u}{v} = (\frac{3}{2})^{\frac{1}{x}} = \frac{-1 \pm \sqrt{7}i}{2}$ .

Hence, we have the ratio of two real numbers equaling one of two non-zero complex numbers $\Rightarrow CONTRADICTION$ . There is no $x \in \mathbb{R}$ that solves the above original exponential equation.

@Corina Margareta Cristian , @Tom Engelsman , how did u solve this question?? Pls share the solution.