<s>[INST] Positive Solution of Exponential Equation 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?
 [/INST] 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.
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