Solve for $n$ in $ \left(n+1 \right) 0.5^n=0.05$

Question

$$ \left(n+1 \right) \times 0.5^n =0.05 $$

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.

Answer 1

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).

Answer 2

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.

Answer 3

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 $7.39$

Answer 4

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 $n$.

Answer 5

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

Answer 6

A direct approach, just numerically approximating, can be done this way (I used Pari/GP).

Reformulating: $$(x+1)/2^x = 0.05 \qquad \qquad \Rightarrow \qquad \qquad x_{k+1} = 0.05 \cdot 2^{x_k} -1 $$

Then I simply iterated , getting the intermediate results, written righthand of the "\":

x = 1             \\  %662 = 1
x = 0.05*2^x1 -1  \\  %663 = -0.900000000000
x = 0.05*2^x1 -1  \\  %664 = -0.973205663437
x = 0.05*2^x1 -1  \\  %665 = -0.974531351027
x = 0.05*2^x1 -1  \\  %666 = -0.974554743334
x = 0.05*2^x1 -1  \\  %667 = -0.974555155908
x = 0.05*2^x1 -1  \\  %668 = -0.974555163184
x = 0.05*2^x1 -1  \\  %669 = -0.974555163312
x = 0.05*2^x1 -1  \\  %670 = -0.974555163315
x = 0.05*2^x1 -1  \\  %671 = -0.974555163315
x = 0.05*2^x1 -1  \\  %672 = -0.974555163315

The second solution -near 8 - occurs, if we reformulate the original formula isolating "the other" x: $$(x+1)/2^x = 0.05 \qquad \qquad \Rightarrow \qquad \qquad x_{k+1} = \log_2 ({1+x_k\over0.05}) $$

x=8  \\  %677 = 8
x=log((1+x)/0.05)/log(2) \\  %678 = 7.49185309633
x=log((1+x)/0.05)/log(2) \\  %679 = 7.40800750864
x=log((1+x)/0.05)/log(2) \\  %680 = 7.39369205256
x=log((1+x)/0.05)/log(2) \\  %681 = 7.39123362954
x=log((1+x)/0.05)/log(2) \\  %682 = 7.39081101764
x=log((1+x)/0.05)/log(2) \\  %683 = 7.39073835663
x=log((1+x)/0.05)/log(2) \\  %684 = 7.39072586343
x=log((1+x)/0.05)/log(2) \\  %685 = 7.39072371536
x=log((1+x)/0.05)/log(2) \\  %686 = 7.39072334602
x=log((1+x)/0.05)/log(2) \\  %687 = 7.39072328252
x=log((1+x)/0.05)/log(2) \\  %688 = 7.39072327160
x=log((1+x)/0.05)/log(2) \\  %689 = 7.39072326972
x=log((1+x)/0.05)/log(2) \\  %690 = 7.39072326940
x=log((1+x)/0.05)/log(2) \\  %691 = 7.39072326934
x=log((1+x)/0.05)/log(2) \\  %692 = 7.39072326933
x=log((1+x)/0.05)/log(2) \\  %693 = 7.39072326933
x=log((1+x)/0.05)/log(2) \\  %694 = 7.39072326933

Answer 7

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.