3

The equation is $2^{x+1}+2^{1/x^2}=6$.

By inspection I see that $1$ is a solution. However, after trying to algebraically isolate for $x$, I was unable to deduce that $1$ is a solution. Given the simplicity of the value of the solution, I was wondering if it would be possible to do so?

Also, I am only looking for the positive solution. However, when graphically analyzing the equation, I noticed that there exists a negative solution that Wolfram Alpha is incapable of giving an exact form for. Does there exist an exact form of the negative solution other than an infinite decimal?

allegro.sostenuto
  • 308

1 Answers1

3

I don't believe it is actually possible to isolate for $x$ in this case, however I was able to "algebraically deduce" that 1 is a solution using the following rationale involving the AM-GM inequality twice: \begin{align*} 2^{x+1}+2^{\frac{1}{x^2}} & = 6 \\ 2(2^x)+2^{\frac{1}{x^2}} & = 6 \\ 2^x+2^x+2^{\frac{1}{x^2}} & = 6 \\ \implies\frac{2^x+2^x+2^{\frac{1}{x^2}}}{3} & \ge \sqrt[3]{2^x2^x2^{\frac{1}{x^2}}} \\ \sqrt[3]{2^{x+x+\frac{1}{x^2}}} & \le 2 \\ x+x+\frac{1}{x^2} & \le 3 \\ \implies\frac{x+x+\frac{1}{x^2}}{3} & \ge \sqrt[3]{1} \\ x+x+\frac{1}{x^2} & \ge 3 \\ \therefore x+x+\frac{1}{x^2} & = 3 \\ x=x & = \frac{1}{x^2} \\ x & = 1 \end{align*} I tried deducing the negative solution. However, I was not able to, nor do I believe it is possible to do so.

Philosophiæ
  • 198