One of the early questions in the NCEES FE Industrial/Systems Exam study guide has a solution that I can't decipher. The question is as follows:
Bob wants to be a millionaire. To achieve this goal, at the end of the year he invests $5000 each year into an account that pays 10% interest, compounded yearly. How many years will it take Bob to reach this goal?
The part I need help with is the solution to this problem, it's written bellow. I have no idea how it gets from $(\$5000)\frac{(1+0.1)^n - 1}{0.1})$ to $21 = 1.1^n$. Can anyone provide an alternate solution or some explanation? Thank you!
\begin{align} A&=\text{uniform series of end of compounding period cash flows}\\ &=\$5000\\ i&=\text{annual interest rate}\\ &=10\%\\ n&=\text{number of compounding periods}\\ &=\text{number of years}\\ F&=\text{future equivalent value of a cash flow or series of cash flows}\\ &=\$1,000,000\\ \end{align}
\begin{align} \$1,000,000&=(\$5000)(F/A,10\%,n)\\ &=(\$5000)\frac{(1+0.1)^n - 1}{0.1})\\ 21&=1.1^n\color{red}{\leftarrow}\\ \log 21 &=n\log 1.1\\ n&=\frac{\log 21}{\log 1.1}=31.94\text{ yr} \end{align}
The $1,000,000 carries down and is equal to the right hand side of the equation in the red box. From there, use the following rearrangements:
$$1000000= 5000\Big(\cfrac{(1+0.1)^n - 1}{0.1}\Big)$$
$$\cfrac{1000000}{5000}= \cfrac{(1+0.1)^n - 1}{0.1}$$
$$\Big(\cfrac{1000000}{5000}\Big)(0.1) = (1 + 0.1)^n - 1$$
$$\Big(\cfrac{1000000}{5000}\Big)(0.1) + 1= 1.1^n$$
$$21 = 1.1^n$$
Hope that helps.