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Suppose $\\var{years}$ years ago you deposited $\\$\\var{P}$ into a bank account and today the balance accumulated is $\\$\\var{S}$. If the interest was compounded {period[0]} what was the rate of compound interest per annum?

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[[0]] $\\%$ (to two decimal places)

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\\var{period[2]}\\left(\\left(\\frac{\\var{S}}{\\var{P}}\\right)^{\\frac{1}{\\var{n}}}-1\\right)

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present value

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We are asked to find the interest rate per annum using compound interest. Therefore we will use the compound interest equation

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$S=P(1+i)^n$,

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where $S$ is the future value, $P$ is the present value, $i$ is the interest rate per time period and $n$ is the number of time periods.

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In our situation we have,

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$S=\\var{S}$

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$P=\\var{P}$,

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$n=\\var{years}$, 

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and therefore we have

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$\\var{S}=\\var{P}\\left(1+r\\right)^\\var{n}$,

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which we need to rearrange to solve for $i$.

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We want to get $i$ by itself. We start by dividing both sides by $\\var{P}$ (to remove the multiplication by $\\var{P}$)

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$\\displaystyle \\frac{\\var{S}}{\\var{P}}=\\left(1+i\\right)^\\var{n}$.

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Next we remove the power on the right hand side by raising both sides to the power of $\\frac{1}{\\var{n}}$ (this is equivalent to applying $\\sqrt[\\var{n}]{\\phantom{XX}}$ to both sides)

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$\\displaystyle\\left(\\frac{\\var{S}}{\\var{P}}\\right)^{\\frac{1}{\\var{n}}}=\\left(\\left(1+i\\right)^\\var{n}\\right)^{\\frac{1}{\\var{n}}}$.

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Which simplifies (by an index law) to 

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$\\displaystyle\\left(\\frac{\\var{S}}{\\var{P}}\\right)^{\\frac{1}{\\var{n}}}=1+i$.

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Next to get $i$ by itself we subtract $1$ from both sides to get 

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$\\displaystyle\\left(\\frac{\\var{S}}{\\var{P}}\\right)^{\\frac{1}{\\var{n}}}-1=i$.

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Recall that $i$ is the interest rate per time period, but in this question this is the same as the interest rate per annum, therefore we have 

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$\\begin{array}\\displaystyle \\text{interest rate pa}&=\\left(\\frac{\\var{S}}{\\var{P}}\\right)^{\\frac{1}{\\var{n}}}-1\\\\&\\approx\\var{ipadec}\\\\&=\\var{iparounded}\\%\\quad \\text{(2 decimal places)}\\end{array}$

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