I just ran out of money by spending $\$1000$ per year on average, while my unspent money was earning 6% per year continuous interest. If I spent $\$13800$ in total, how much did I have when I started spending, and how long has it been since then?
I haven't found any posts that explain how to invert the formula, especially for the combination of continuous interest and continuous spending. (Thanks!)
Maybe you can just reverse time on this problem and use the standard equation to solve it, something like:
I have just started with \$0 in my bank account. I deposit exactly \$1000 every year, but unfortunately the amount in the account depreciates continuously at a rate of 6% annually. By the time I've put in a total of \$13800 into the account, how much time has passed, and how much money will I actually have in my account by then?
That being said, you can also work backwards with the information you have.
If you spent \$13800 in total, and the only thing you spent money on was exactly \$1000 per year, then you must have taken 13.8 years to run out of money.
Given this, you can use the standard formula for computing compound interest and expenses in order to find your initial amount of money: you know the interest rate, the spending rate, the total time, and the final amount of money (\$0).
I'm not sure of the standard formula, but I believe it must be something like $$P(t+1) = P(t) \cdot 1.06 - 1000$$
which, if you work it out, gives the general formula: $$P(t) = P_0 \cdot 1.06^t - 1000\cdot \sum_{k=0}^{k=t-1} 1.06^k$$
or just $$P(t) = P_0 \cdot 1.06^t - 1000\cdot \frac{1.06^{t}-1}{0.06}$$
where $P(t)$ is the amount of money you have over time, $P_0$ is your initial amount of money, and $t$ is the amount of time since the initial investment, measured in years. I'm assuming that interest is compounded continuously, as you suggest, and similarly that expenses accrue gradually over the course of the year, rather than all at once. (If your formula differs, the solution process is still the same.)
You run out of money after 13.8 years, so: $$0 = P_0 \cdot 1.06^{13.8} - 1000\cdot \frac{1.06^{13.8}-1}{0.06}$$ $$P_0 = 1000 \cdot \frac{1.06^{13.8}-1}{0.06}\cdot \frac{1}{1.06^{13.8}} $$ $$P_0 = 1000 \cdot \frac{1-1.06^{-13.8}}{0.06}$$ $$P_0 \approx 9208.57$$
If we plug this into our original formula, we find that it checks out at time $t=13.8$.
$800\cdot(1.06)^{-14} + 1000\sum_\limits{n=1}^{13} (1.06)^{-n} = B\\ 800\cdot(1.06)^{-14} + 1000 \frac {(1-1.06^{-13})}{0.06} = B\\ B = 9,206.5$
