Using Newton-Raphson method, find the solution for $e^{\frac{x^2}{4vt}} = 1+\frac{x^2}{2vt}$ I need help with solving this difficult fluid dynamic expression. I have tried using rules of logs, symbolab algebra calculator and Wolfram Alpha calculator, and I have got no solution.

How would you solve the following expression USING the NEWTON-RAPHSON method for $x$?
  $$e^{\frac{x^2}{4vt}} = 1+\frac{x^2}{2vt}$$

When solving this USING the NEWTON-RAPHSON method, the solution is: $x=2.2418\sqrt{vt}$

I want to know how you could solve the first expression using the NEWTON-RAPHSON method to get the solution. So could someone please provide a step-by-step solution, by using this method please? 

Note: This question was answered, however it was NOT answered using NEWTON-RAPHSON method. It was answered using the Lambert W function, which is a very long and complicated process as compared to the Newton-Raphson method. 
 A: Well, we have:
$$\exp\left(\frac{x^2}{4\cdot\text{v}\cdot t}\right)=1+\frac{x^2}{2\cdot\text{v}\cdot t}\tag1$$
Now, we know that we can write:
$$\exp\left(\alpha\right)=\sum_{\text{n}=0}^\infty\frac{\alpha^\text{n}}{\text{n}!}=\frac{\alpha^0}{0!}+\frac{\alpha^1}{1!}+\frac{\alpha^2}{2!}+\dots=$$
$$1+\alpha+\frac{\alpha^2}{2}+\dots\tag2$$
So, for equation $(1)$ we can write:
$$1+\frac{x^2}{4\cdot\text{v}\cdot t}+\frac{1}{2}\cdot\left(\frac{x^2}{4\cdot\text{v}\cdot t}\right)^2+\dots=1+\frac{x^2}{2\cdot\text{v}\cdot t}\tag3$$
Using the aproximation of three terms we have:
$$1+\frac{x^2}{4\cdot\text{v}\cdot t}+\frac{1}{2}\cdot\left(\frac{x^2}{4\cdot\text{v}\cdot t}\right)^2\approx1+\frac{x^2}{2\cdot\text{v}\cdot t}\space\Longleftrightarrow\space$$
$$x\approx0\space\vee\space x\approx\pm2\sqrt{2}\cdot\sqrt{\text{v}\cdot\text{t}}\tag4$$
A: First express this in terms of a single variable: letting $s = x/\sqrt{vt}$, the equation becomes 
$$ e^{s^2/4} = 1 + s^2/2$$
Now with $f(s) = \exp(s^2/4) - (1 + s^2/2)$, $f'(s) = s \exp(s^2/4)/2 - s$, and the Newton iteration is 
$$ s_{n+1} = s_n - \frac{f(s_n)}{f'(s_n)}$$ 
Note that $s=0$ is also a solution, so you don't want to start too close to that. 
Starting with, say, $s_0 = 2$, you just iterate until the numbers get close enough to each other.
$s_1 = 2 - f(2)/f'(2) = 2.392211192$
$s_2 = 2.392211192 - f(2.392211192)/f'(2.392211192) = 2.269512712$
etc.
I find that $s_5$ and $s_6$ differ only in the $9$'th decimal place.
A: We put $0 \le y=x^2/(4vt)$ and look for the non-negative zeros of the function $f(y)$
$$
\left\{ \matrix{
  f(y) = e^{\,y}  - 2y - 1 = 0 \hfill \cr 
  f'(y) = e^{\,y}  - 2\quad  \Rightarrow \quad \min f(y):\;y = \ln 2 \hfill \cr 
  0 < f''(y) = e^{\,y}  \hfill \cr}  \right.
$$
Clearly, $f(y)$ is convex, has a negative minimum at $y=ln2$, thus it has two zeros.
One of them is at $y=0$ and the other will be past the minimum.
Since for $ln2 < y$ the function is increasing we can apply Newton-Raphson
method to find the second zero, provided that the starting point $y_0$ be to the right of the minimum.

We can choose $y_0=2$ for instance, and then start the recursion
$$
\eqalign{
  & {{f(y_0 )} \over {y_1  - y_0 }} = f'(y_0 )\quad  \Rightarrow \quad   \cr 
  &  \Rightarrow \quad y_1  = y_0  + {{f(y_0 )} \over {f'(y_0 )}} = y_0  + {{e^{\,y_0 }  - 2y_0  - 1} \over {e^{\,y_0 }  - 2}}\quad  \Rightarrow   \cr 
  &  \Rightarrow \quad y_{n + 1}  = y_n  + {{e^{\,y_{\,n} }  - 2y_n  - 1} \over {e^{\,y_{\,n} }  - 2}} \cr} 
$$
Of course, once found a satisfactory value for $y$, you can easily 
convert it back to $x$
A: Just as Will Jagy did, let$u = \frac{x^2}{4 v t}$ to make the equation $e^u - 1 - 2u = 0$.
So, let consider that you look for the zero's of function
$$f(u)=e^u - 1 - 2u $$ for which 
$$f'(u)=e^u - 2 \qquad \text{and} \qquad f''(u)=e^u > \,\,\forall u$$
The first derivative cancels when $u=\log(2)$. You can get an estimate of the root builiding the Taylor series at this point. This would give
$$e^u - 1 - 2u =(1-2 \log (2))+(u-\log (2))^2+O\left((u-\log (2))^3\right)$$ Ignoring the gigher order terms, you then have as an estimate
$$u_0=\log(2)+\sqrt{2 \log (2)-1}\approx 1.31467$$ With this estimate, you can now use Newton method
$$u_{n+1}=u_n-\frac{f(u_n)}{f'(u_n)}=\frac{e^{u_n} (u_n-1)+1}{e^{u_n}-2}$$ and get, for  twelve  significant figures, the following iterates
$$\left(
\begin{array}{cc}
 n & u_n \\
 0 & 1.31467301359 \\
 1 & 1.26002526328 \\
 2 & 1.25644611685 \\
 3 & 1.25643120888 \\
 4 & 1.25643120863
\end{array}
\right)$$
One of the important points when you use Newton method is to get a "reasonable" estimates.
You will notice in the table that, at no time, we overshoot the solution because we started at a point whe $f(x_0) \times f''(x_0) > 0$ (Darboux theorem).
Starting instead with $x_0=1$, the iterates would have been
$$\left(
\begin{array}{cc}
 n & u_n \\
 0 & 1.00000000000 \\
 1 & \color{red}{1.39221119118} \\
 2 & 1.27395717022 \\
 3 & 1.25677778598 \\
 4 & 1.25643134800 \\
 5 & 1.25643120863 
\end{array}
\right)$$
