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
You seem to want the details of a numerical method to find an good approximation for $x$. Here is one way, using the Newton-Raphson method, which is well-known and uses fewer steps than other famous methods.

We want to find the zeros of

$$f(x)=0.25^x+0.5^x+0.75^x-1$$

(I rearranged the terms to be in ascending order, which satisfies my sense of style. I also changed the fractions to decimals to make it easier to type into a calculator.) We already know from comments and other answers that $x=1$ is too low and $x=2$ is too high. So let's choose the average, $1.5$, as our initial guess $x_0$. We then use the recurrence 

$$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$

This is what I typed into my graphing calculator (TI-Nspire CX):

$$x:=1.5$$
$$x:=x-\frac{0.25^x+0.5^x+0.75^x-1}{0.25^x\cdot\ln(0.25)+0.5^x\cdot\ln(0.5)+0.75^x\cdot\ln(0.75)}$$

That gave me the answer
$$1.71161782284$$

which is $x_1$. Repeated presses of the [enter] key repeated the recurrence, and I got

$$1.73037816962$$
$$1.73050735181$$
$$1.73050735786$$
$$1.73050735786$$

More [enter]'s just repeated the last value. Therefore, an excellent approximation to your desired value, probably to $12$ significant digits, is $1.73050735786$.
$\left ( \frac{1}{2} \right )^{n}+\left ( \frac{1}{4} \right )^{n}+\left ( \frac{3}{4} \right )^{n}= \frac{2^n}{4^n} +\frac{1^n}{4^n} +\frac{3^n}{4^n} = \frac{1^n+2^n+3^n}{4^n}=1,\: \therefore 1^n+2^n+3^n=4^n  $  

I don't think this can be solved analytically,
you just have to use trial and error with n