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. 

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.
