Solve for $n$ in $ \left(n+1 \right) 0.5^n=0.05$ $$ \left(n+1 \right) \times 0.5^n =0.05 $$
Is there a way to solve this directly for $n$? I know that by taking logs we can simplify it but we still do not get a value as far as I can see. A solution is to start plugging numbers starting from 1 and continuing up to 8 which seems to be the closest approximation. Any hints? Thanks.
 A: You might coerce a solution by using the Lambert W function, but that is not usually considered elementary.  For a numeric solution, I would take base 2 logs of both sides: $\log_2(n+1)-n=\log_2 0.05$ and render it as $\log_2(n+1)-\log_20.05=n$  Since the log is slowly varying, I would set the $n$ on the left to zero, calculate the left side, plug that $n$ in and iterate.  It converges  rapidly to about $7.39$
A: There are ways to solve for $n$ in this equation using polylogarithms--but this is very much nontrivial and probably not what you are looking for. Most likely, this is something you would do graphically or with a computer algebra program, such as Mathematica or wolframalpha.
EDIT. As you requested, I approximated solutions using Mathematica and obtained $n\approx -0.9745551633147583$ and $n\approx 7.390723269331218$. (Although Mathematica can also produce exact results).
A: I will propose a more elementary solution. 
Rewrite the equation as $20(n+1)=2^n$.  A simple plot will show you there are two solutions.  Assuming you're looking for the positive you can easily bracket the solution between 7 and 8.
A: For this class of equations, what you suggest is good : try to bracket if you know that there is only one root. Another solution is to plt the function to see and locate the possible roots. Then, using a guess, apply Newton method up to the desired accuracy. In your case, assuming "n" real, there are two roots at -0.974 and at 7.391. Think about Wolfram Alpha.
A: Denote $v=n+1$
$$
(n+1)5^n=0.05\\
\frac{(n+1)5^{n+1}}{5}=0.05\\
(n+1)5^{n+1}=0.25\\
ve^{v \log 5}=0.25
$$
Now you have Lambert-W function that you can solve for $v$. Then replace $v$ with $n+1$ to obtain the value for $n$.
A: Using Lambert W we get as a first solution, all steps are displayed:
$$ \begin{eqnarray} (x+1) \cdot 2^{-x} &=& 5/100=1/20 \\ 
(x+1) \cdot 2^{-(x+1)} &=& 1/40 \\
-(x+1) \cdot 2^{-(x+1)} &=& -1/40 \\ 
-(x+1)\cdot e^{-\ln2 \cdot (x+1)} &=& -1/40 \\ 
-\ln 2\cdot  (x+1)\cdot e^{-\ln2 \cdot (x+1)} &=& -{ \ln 2 \over 40} \\ 
 y e^y &=& - { \ln 2 \over 40} \\
 y &=& W(- { \ln 2 \over 40}) \\
 x &=& {y \over -\ln 2} - 1 \\
 x &\approx& -0.974555163315  \qquad \qquad \text{ first solution }
 \end{eqnarray}
$$
Using different branches of the Lambert W we should get the other solutions as well
