Use the given rate of inflation to calculate the value of $100 in 1910, now. problem:
You have been told that the average rate of inflation between 1910 and 2016 is $3.8\%$. Using this information, calculate how much $100 in 1910 would be worth today.
What I tried:
This one is a little tough because I know inflation is compounded. I know of this formula:
$$ inflation = \frac {final - initial}{initial}$$
Except this formula doesn't consider the fact that inflation would compound upon itself. Do I need to modify the above formula or is there another I can use?
 A: You wish to compute the number of 2016 dollars that are equivalent to $\$100$ 1910 dollars given an average inflation rate of $3.8\%$ per year.  
For our purposes, it suffices to treat the inflation rate as constant.  Then if $v_i$ is the amount at the beginning of the year, $v_f$ is the amount at the end of the year, and $100r\%$ is the inflation rate, your formula yields
$$100r\% = \frac{v_f - v_i}{v_i} \cdot 100\%$$
Solving for $v_f$ yields 
\begin{align*}
r & = \frac{v_f - v_i}{v_i}\\
rv_i & = v_f - v_i\\
rv_i + v_i & = v_f\\
v_i(1 + r) & = v_f
\end{align*}
Let $v_0$ be the initial amount. Let $v(t)$ be the amount after $t$ years.  The calculation we did above shows that 
$$v(1) = v_0(1 + r)$$
Iterating, we find that 
\begin{align*}
v(2) & = v(1)(1 + r)\\
     & = [v_0(1 + r)](1 + r)\\ 
     & = v_0(1 + r)^2\\
v(3) & = v(2)(1 + r)\\
     & = [v_0(1 + r)^2](1 + r)\\
     & = v_0(1 + r)^3\\
v(4) & = v(3)(1 + r)\\
     & = [v_0(1 + r)^3](1 + r)\\
     & = v_0(1 + r)^4
\end{align*}
which suggests that $$v(t) = v_0(1 + r)^t$$ which can be proved by mathematical induction.  
In your example, $v_0 = \$100$, $r = \frac{3.8\%}{100\%} = 0.038$, and $t = 2016 - 1910 = 116$.
A: Inflation represents the increase in the price of consumer products over a period of time. If $i=3.8\%$ represents he average rate of inflation between $t_i=1910$ and $t_f=2016$, then the value $P$ at time $t_f$
$$
P(t_f)=P(t_i)(1+i)^{t_f-t_i}=100\times(1+3.8\%)^{106}= 2,491.65 
$$
A: Solve for $x$
$$x\bigg(1+\frac{3.8}{100}\bigg)^{2016-1910} = 100$$
