How do you solve for x in this equation? I tried expanding, but I still can't get rid of the exponents to isolate x. 
$$\frac{(1+x)^4-1}{x}=4.374616$$
Thank you in advance for your help. 
 A: You want to solve $$\frac{(1+x)^4-1}{x}=a$$ After expansion and simplification, this equation write $$x^3+4x^2+6x=a-4$$ you could solve using Cardano's formula. On the other side, you can consider that the solution is the intersection of the function $$y=x^3+4x^2+6x$$ with the horizontal line $y=a-4$. If you study the function, you can prove that it does not show any maximum since its derivative $$y'=3x^2+8x+6$$ has no real root. So, there is only a real solution which will be positive if $a \gt 4$, $0$ if $x=4$ and negative if $a \lt 4$.
Since in your case, $a$ is not much larger than $4$, the solution is close to $0$ and a Newton procedure can be used. Starting with an initial guess $x_0=0$, Newton will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ Using $f(x)=x^3+4x^2+6x-0.374616$, the successive iterates will be $0.062436$, $0.0600038$, $0.0600000$ which is the exact solution.
A: \begin{align}
(1+x)^{4} - 1 &= [(1+x)^{2} - 1] [ (1+x)^{2} + 1] = (1+x-1)(1+x+1)(x^{2} + 2x +1)
\nonumber\\
&= x(x+2)(x^{2}+2x+2)
\end{align}
or
\begin{align}
\frac{(1+x)^{4} -1}{x} = (x+2) (x^{2} + 2x +2).
\end{align}
Now, $4.374616$ can be factored into 
\begin{align}
4.374616 = (2.06)(2.1236) = (2 + .06)((.06)^{2} + 2(.06) + 2).
\end{align}
This yields
\begin{align}
(x+2) (x^{2} + 2x +2) = (2 + .06)((.06)^{2} + 2(.06) + 2)
\end{align}
which gives $x = .06$. 
A: I will present two general approaches, starting with the one that works best for this particular case:

Assuming small values of x, we have $(1+x)^4\approx1+4x+6x^2$. Subtracting $1$ and dividing by x, we are left with solving $4+6x=4+\alpha\iff x=\dfrac\alpha4=0.06,~$ where $\alpha=0.374616$.

Let $~f(x)=(1+x)^4.\quad$ Then $~\dfrac{(1+x)^4-1}x=\dfrac{(1+x)^4-(1+0)^4}{x-0}\approx f'(0).\quad$ But $~f'(x)=$ 
$=4\cdot(1+x)^3.\quad$ So $~4\cdot(1+x)^3\approx a\iff x\approx-1+\sqrt[3]{\dfrac a4}=0.03,~$ which unfortunately is 
only half of the true value, $x=0.06.~$ Nevertheless, the reason I chose to present this method as well is because, despite not working very well for this particular case, it works quite well in many other similar situations.
