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

The function :
$$y=0.25^x+0.5^x+0.75^x-1$$
is decreasing. 
For example $y(1)=2$ and $y(2)=-\frac{1}{8}$. So, the root for $y=0$ is  between $x=1$ and $x=2$.
In this case, among many numerical methods, the dichotomic method is very simple. The successives values $x_k$ are :
$$x_{k+1}=x_{k}+\frac{\delta_k}{2^k}$$
where $\delta_k=\pm 1$
the signe is $+$ if $y_k=(0.25^{x_k}+0.5^{x_k}+0.75^{x_k}-1) >0$ 
the signe is $-$ if $y_k<0$.
ALGORITHM:
$x:=1$
$d:=1$
repeat 
$ \quad \quad d:=\frac{d}{2}$
$ \quad \quad y:=0.25^x+0.5^x+0.75^x-1$
$ \quad \quad $ if $y>0$ then $x:=x+d$ else $x:=x-d$
until $d<10^{-15} $ 
(or another limit depending on the wanted accuracy).
RESULT : $x=1.73050735785763$
Of course, the convergence is slower than with the Newton-Raphson method for example. But, in both casses, the time of computation is so small that this is negligible. On the other hand, the time spent in programming the algorithm is smaller with the dichotomic method : that is the most important point in practice. 
