Exponential problem

Question

$\$10,000$ increases every day by $1\%$. How long until it doubles? I tried doing it by multiplying by $1\%$ for each day and got $10$ days but I don't think that is right. I know there must be an easier way to do it.

Answer 1

Find a formula that tells you how much money you have after $n$ days. For example:

After $0$ days, you have $a_0=10,000$.

After $1$ day, you have $a_1=10,000 + 0.01\cdot 10,000 = 10,000 \cdot 1.01$

After $2$ days, you have $a_2 = a_1 + 0.01\cdot a_1 = 1.01a_1 = 10,000\cdot 1.01\cdot 1.01=10.000\cdot 1.01^2$...

Can you see a pattern?

Answer 2

How about this: you want to find the value of $t$ so that : $$ (1+ \frac {1}{100})^t=2$$

This follows from the fact that every day your money increases by $1$% .Day $1$ you have $10,000$. Day $2$ , you have $10,000+ \frac{1}{100}\times 10,000=10,000(1+ \frac {1}{100}) $ (equality follows by factoring $10,000$ from both terms ). Every day, the amount of money is multiplied by this same amount $(1+ \frac {1}{100})$

Then you can isolate (and solve for ) $t$ by taking $ln$ on both sides:

I'm not sure how your calculator works, but maybe you can use the identity $$(1+\frac{1}{100})^t=e^{t.ln(1.01)} $$ as a function of $t$, after finding the numerical value for $ln(1.01)$, plug in values of $t$. Does this work?

Answer 3

Not marked as homework, so...

For day $1$, multiply $10\ 000$ by $1.01$ ($101\%$ of $10\ 000$). Continue for each day like this:

$\text{day }0:\ \ 10\ 000\\ \text{day }1:\ \ 10\ 000\times1.01\\ \text{day }2:\ \ 10\ 000\times1.01\times1.01\\ \text{day }3:\ \ 10\ 000\times1.01\times1.01\times1.01\\ \text{day }4:\ \ 10\ 000\times1.01\times1.01\times1.01\times1.01\\ \ \ \ \vdots\\ $

You could continue like this on a calculator till you reached $20\ 000$. Alternatively ...

For day $n$, write $10\ 000\times1.01^n$.

$1.01^n\times10\ 000=20\ 000\\ 1.01^n=\dfrac{20\ 000}{10\ 000}\\ 1.01^n=2$

Since $a^n=b$ can be rewritten $n=\dfrac{\ln(b)}{\ln(a)},$

$ n=\dfrac{\ln(2)}{\ln(1.01)}\text{ days} $

For further information, see here.