1.) What is 3,200,000,000 in scientific notation?
2.) Assume you have $1,000 and a place to invest it that will give you 5% compounded annually. How long will it take your investment to be worth $1,500? Is it worth your while to run around and find another place which will give you 5% interest compounded daily?
+118724 Annual compounding
$$FV=PV(1+i)^n$$
$$\\1500=1000(1+0.05)^n\\
3=2(1.05)^n\\
1.5=1.05^n\\
log1.5=log1.05^n\\
log1.5=nlog1.05\\
n=\frac{log1.5}{log1.05}\\$$
$${\frac{{log}_{10}\left({\mathtt{1.5}}\right)}{{log}_{10}\left({\mathtt{1.05}}\right)}} = {\mathtt{8.310\: \!386\: \!222\: \!520\: \!560\: \!2}}$$
8.31 years
Daily compounding
$$\\1500=1000(1+0.05/365)^{365n}\\
1.5=(1+0.05/365)^{365n}\\
log1.5=log(1+0.05/365)^{365n}\\
log1.5=365n*log(1+0.05/365)\\
365n=\frac{log1.5}{log(1+0.05/365)}\\
n=\frac{log1.5}{365log(1+0.05/365)}\\$$
$${\frac{{log}_{10}\left({\mathtt{1.5}}\right)}{\left({\mathtt{365}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.05}}}{{\mathtt{365}}}}\right)\right)}} = {\mathtt{8.109\: \!857\: \!581\: \!140\: \!344\: \!3}}$$
8.11 years (2dp)
The difference if 0.2 of a year = 2.4 months
Well if you look around for more than 2.4 months you will start losong money.
BUT If the interest is only added yearly with the first one then you might have to wait for the full year 9 years before you can withdraw the $1500.
With the daily one I would assume that the interest is added to the account daily and you can withdraw the $1500 in 8.11 years.
Now the daily one is looking quite a lot better.
+118724 Annual compounding
$$FV=PV(1+i)^n$$
$$\\1500=1000(1+0.05)^n\\
3=2(1.05)^n\\
1.5=1.05^n\\
log1.5=log1.05^n\\
log1.5=nlog1.05\\
n=\frac{log1.5}{log1.05}\\$$
$${\frac{{log}_{10}\left({\mathtt{1.5}}\right)}{{log}_{10}\left({\mathtt{1.05}}\right)}} = {\mathtt{8.310\: \!386\: \!222\: \!520\: \!560\: \!2}}$$
8.31 years
Daily compounding
$$\\1500=1000(1+0.05/365)^{365n}\\
1.5=(1+0.05/365)^{365n}\\
log1.5=log(1+0.05/365)^{365n}\\
log1.5=365n*log(1+0.05/365)\\
365n=\frac{log1.5}{log(1+0.05/365)}\\
n=\frac{log1.5}{365log(1+0.05/365)}\\$$
$${\frac{{log}_{10}\left({\mathtt{1.5}}\right)}{\left({\mathtt{365}}{\mathtt{\,\times\,}}{log}_{10}\left({\mathtt{1}}{\mathtt{\,\small\textbf+\,}}{\frac{{\mathtt{0.05}}}{{\mathtt{365}}}}\right)\right)}} = {\mathtt{8.109\: \!857\: \!581\: \!140\: \!344\: \!3}}$$
8.11 years (2dp)
The difference if 0.2 of a year = 2.4 months
Well if you look around for more than 2.4 months you will start losong money.
BUT If the interest is only added yearly with the first one then you might have to wait for the full year 9 years before you can withdraw the $1500.
With the daily one I would assume that the interest is added to the account daily and you can withdraw the $1500 in 8.11 years.
Now the daily one is looking quite a lot better.
