<s>[INST] Solving for n in the equation $\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$ 
Solving for $n$ in the equation $$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}=1$$

Can anyone show me a numerical method step-by-step to solve this?
Thanks
 [/INST] Classic problem for Newton's Method.
Note that $F(n) = \dfrac{1}{2^n} + \dfrac{1}{4^n} + \dfrac{3^n}{4^n}-1$ is continuously differentiable.  Furthermore, $F(0) = 2$ and $\lim_{n \rightarrow \infty} F(n) =-1$ so by the Intermediate Value Theorem, there is definitely a zero.
I used Maple to calculate the root to a tolerance of $10^{-15}$ because I can.
with(Student[NumericalAnalysis]):
F(z):= n-> 1/2^n + 1/4^n + 3^n/4^n:


Newton(F(z),n=0.2,tolerance=10^-15)
>>>1.73050735785764

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