Q: Solve the following equation $ \frac{1}{2^x} + \frac{1}{4^x} + \frac{1}{6^x} + \frac{1}{8^x} = 1 $ . How can I solve the following equation? 

$$ 
\dfrac{1}{2^x} + 
\dfrac{1}{4^x} + 
\dfrac{1}{6^x} + 
\dfrac{1}{8^x} = 
1 
$$ 


A: Write the equation as $2^{-x} + 4^{-x} + 6^{-x} + 8^{-x} = 1$. Define the function $g(x) = 2^{-x} + 4^{-x} + 6^{-x} + 8^{-x}$. Then, $g$ is a differentiable function, with $g'(x) = -2^{-x}\ln 2 - 4^{-x}\ln 4 - 6^{-x}\ln 6 - 8^{-x}\ln 8$, which is always negative. Hence, $g$ is a decreasing continuous function, so it's a one-one map. That is, there exists a single root of the above equation.
Furthermore, since $g(1) = \frac{25}{24} > 1$, and $g(2) = \frac{205}{576} < 1$, we have by the intermediate value theorem, that the root is between $1$ and $2$. Also, from the closeness of $\frac{25}{24}$ to $1$, we would be inclined to think that such a root is very close to $1$. Indeed, as far as Wolfram Alpha is concerned, the root is $1.03411$, and from our analysis, we confirm this as being the unique root.
I do not think there are purely algebraic ways of solving this problem (our method involved calculus). 
