Nzuzo invests \(\text{R}\,\text{80 000}\) at an interest rate of \(\text{7,5}\%\) per annum compounded yearly. How long will it take for his investment to grow to \(\text{R}\,\text{100 000}\)?
It will take just over \(\text{3}\) years.
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2.8 Enrichment: more on logarithms

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3.2 Annuities

In earlier grades we studied simple interest and compound interest, together with the concept of depreciation. Nominal and effective interest rates were also described.
In this chapter we look at different types of annuities, sinking funds and pyramid schemes. We also look at how to critically analyse investment and loan options and how to make informed financial decisions.
Financial planning is very important and it allows people to achieve certain goals, such as supporting a family, attending university, buying a house, and saving enough money for retirement. Prudent financial planning includes making a budget, opening a savings account, wisely investing savings and planning for retirement.
For calculations using the simple interest formula, we solve for \(n\), the time period of an investment or loan, by simply rearranging the formula to make \(n\) the subject. For compound interest calculations, where \(n\) is an exponent in the formula, we need to use our knowledge of logarithms to determine the value of \(n\).
\[A = P{\left(1+i\right)}^{n}\] \begin{align*} A &= \text{accumulated amount} \\ P &= \text{principal amount} \\ i &= \text{interest rate written as a decimal} \\ n &= \text{time period} \end{align*}Solving for \(n\):
\begin{align*} A & = P{\left(1+i\right)}^{n} \\ \frac{A}{P} & = {\left(1+i\right)}^{n} \\ \text{Use definition: } n & = \log_ {\left(1+i\right)}{\left( \frac{A}{P} \right)} \\ \text{Change of base: } n &= \frac{\log\left(\frac{A}{P}\right)}{\log(1 + i)} \end{align*}Thembile invests \(\text{R}\,\text{3 500}\) into a savings account which pays \(\text{7,5}\%\) per annum compounded yearly. After an unknown period of time his account is worth \(\text{R}\,\text{4 044,69}\). For how long did Thembile invest his money?
The \(\text{R}\,\text{3 500}\) was invested for \(\text{2}\) years.
Margo has \(\text{R}\,\text{12 000}\) to invest and needs the money to grow to at least \(\text{R}\,\text{30 000}\) to pay for her daughter's studies. If it is invested at a compound interest rate of \(\text{9}\%\) per annum, determine how long (in full years) her money must be invested?
In this case we round up, because \(\text{10}\) years will not yet deliver the required \(\text{R}\,\text{30 000}\). Therefore the money must be invested for at least \(\text{11}\) years.
Nzuzo invests \(\text{R}\,\text{80 000}\) at an interest rate of \(\text{7,5}\%\) per annum compounded yearly. How long will it take for his investment to grow to \(\text{R}\,\text{100 000}\)?
It will take just over \(\text{3}\) years.
Sally invests \(\text{R}\,\text{120 000}\) at an interest rate of \(\text{12}\%\) per annum compounded quarterly. How long will it take for her investment to double?
It will take just over \(\text{5}\) years and \(\text{10}\) months.
When Banele was still in high school he deposited \(\text{R}\,\text{2 250}\) into a savings account with an interest rate of \(\text{6,99}\%\) per annum compounded yearly. How long ago did Banele open the account if the balance is now \(\text{R}\,\text{2 882,53}\)? Write the answer as a combination of years and months.
This is a compound interest problem:
\begin{align*} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{2 882,53} \\ P & = \text{2 250} \\ i & = \text{0,0699} \end{align*}Now substitute the known values and solve for \(n\):
\begin{align*} \text{2 882,53} & = \text{2 250} \left( 1 + \text{0,0699} \right) ^ {n} \\ \text{2 882,53} & = \text{2 250}(\text{1,0699})^{n} \\ \frac{\text{2 882,53}}{\text{2 250}} & = (\text{1,0699})^{n} \end{align*}Change to logarithmic form:
\begin{align*} & n = \log_{\text{1,0699}} \left(\frac{\text{2 882,53}}{\text{2 250}} \right) \\ & n = \text{3,6666} \ldots \end{align*}Banele left the money in the account for about \(\text{3,67}\) years.
However, we must give our answer in terms of years and months, not as a decimal number of years. \(\text{3,6666} \ldots\) years means \(\text{3}\) years and some number of months; to figure out how many months, we need to convert \(\text{0,6666} \ldots\) years into months.
We know that there are \(\text{12}\) months in a year.
To convert \(\text{0,6666} \ldots\) years into months, do the following:
\[(\text{0,6666} \ldots )\text{ year} \times \left( \frac{12\text{ months}}{\text{year}} \right) = \text{8} \text{ months}\]Banele deposited the money into the account \(\text{3}\) years and \(\text{8}\) months ago.
The annual rate of depreciation of a vehicle is \(\text{15}\%\). A new vehicle costs \(\text{R}\,\text{122 000}\). After how many years will the vehicle be worth less than \(\text{R}\,\text{40 000}\)?
The vehicle will be worth less than \(\text{R}\,\text{40 000}\) after about \(\text{7}\) years.
Some time ago, a man opened a savings account at KMT South Bank and deposited an amount of \(\text{R}\,\text{2 100}\). The balance of his account is now \(\text{R}\,\text{3 160,59}\). If the account gets \(\text{8,52}\%\) compound interest p.a., determine how many years ago the man made the deposit.
We write the compound interest formula and the given information:
\begin{align*} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{3 160,59} \\ P & = \text{2 100} \\ i & = \text{0,0852} \end{align*}We know everything except for the value of \(n\). Substitute and then solve for \(n\).
\begin{align*} \text{3 160,59} & = \text{2 100} \left( 1 + \text{0,0852} \right) ^ {n} \\ \text{3 160,59} & = \text{2 100}(\text{1,0852})^{n} \\ \frac{\text{3 160,59}}{\text{2 100}} & = (\text{1,0852})^{n} \end{align*}Use the definition of a logarithm to solve for \(n\):
\[n = \log_{\text{1,0852}} \left(\frac{\text{3 160,59}}{\text{2 100}} \right)\]Use a calculator to evaluate the \(\log\):
\[n = \text{4,999} \ldots\]The man made the deposit \(\text{5}\) years ago.
It will take \(\text{11}\) years for their deposit to double in value.
A university lecturer retires at the age of \(\text{60}\). She has saved \(\text{R}\,\text{300 000}\) over the years.
If she manages her money carefully, she will be \(\text{71}\) years or older.
It will take less than \(\text{5}\) years.
Simosethu puts \(\text{R}\,\text{450}\) into a bank account at the Bank of Upington. Simosethu's account pays interest at a rate of \(\text{7,11}\%\) p.a. compounded monthly. After how many years will the bank account have a balance of \(\text{R}\,\text{619,09}\)?
Use the compound interest formula and determine the value of \(n\).
\begin{align*} A & = P(1+i)^n \\ \text{Where: } & \\ A & = \text{619,09} \\ P & = \text{450} \\ i & = \text{0,0711} \end{align*}In this question the interest is payable every month. Therefore \(i \rightarrow \frac{\text{0,0711}}{\text{12}}\) and \(n \rightarrow (n \times \text{12})\). In this case, \(n\) represents the number of years; the product \((n \times \text{12})\) represents the number of times the bank pays interest into the account.
\begin{align*} \text{619,09} & = \text{450} \left( 1 + \frac{\text{0,0711}}{\text{12}} \right) ^ {(n \times \text{12})} \end{align*} \begin{align*} \text{619,09} & = \text{450}(\text{1,00592} \ldots)^{\text{12}n} \\ \frac{\text{619,09}}{\text{450}} & = (\text{1,00592} \ldots)^{\text{12}n} \\ \text{1,37575} \ldots & = (\text{1,00592} \ldots )^{\text{12}n} \end{align*}At this point we must change the equation to logarithmic form:
\begin{align*} \text{12}n & = \log_{\text{1,005925}} (\text{1,37575} \ldots ) \\ \text{12}n & = \text{54} \end{align*}To find the number of years, we solve for \(n\):
\begin{align*} \text{12}n & = \text{54} \\ n & = \frac{\text{54}}{\text{12}} \\ n & = \text{4,5} \end{align*}The money has been in the Simosethu's account for \(\text{4,5}\) years.
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2.8 Enrichment: more on logarithms

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3.2 Annuities
