Suppose you borrow $\\$\\var{P}$ to buy a house and you make {period[0]} repayments of $\\$\\var{C}$ at the beginning of each {period[2]}. Interest is $\\var{ipa}\\%$ per annum compounding {period[0]}. Calculate how many {period[2]+'s'} it will take to pay off the loan.
\n\n[[0]] {period[2]+'s'}
", "useCustomName": false, "gaps": [{"maxValue": "nrounded", "variableReplacements": [], "allowFractions": false, "marks": 1, "mustBeReducedPC": 0, "minValue": "nrounded", "useCustomName": false, "showFeedbackIcon": true, "showCorrectAnswer": true, "unitTests": [], "customName": "", "customMarkingAlgorithm": "", "correctAnswerFraction": false, "correctAnswerStyle": "plain", "extendBaseMarkingAlgorithm": true, "mustBeReduced": false, "type": "numberentry", "variableReplacementStrategy": "originalfirst", "scripts": {}, "showFractionHint": true, "notationStyles": ["plain", "en", "si-en"]}], "showFeedbackIcon": true, "showCorrectAnswer": true, "unitTests": [], "customName": "", "customMarkingAlgorithm": "", "extendBaseMarkingAlgorithm": true, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "scripts": {}, "sortAnswers": false}], "variable_groups": [], "extensions": [], "advice": "You are asked to find the number of time periods of an annuity due (since the payments are at the beginning of each period) and we are given the present value of the annuity. Therefore we will use the present value of an annuity due formula
\n$\\displaystyle P=\\frac{C}{i}\\left(1-\\frac{1}{(1+i)^n}\\right)(1+i)$
\nwhere $P$ is the present value, $C$ is the cash flow per period, $i$ is the interest rate per period, and $n$ is the number of periods.
\nIn our situation we have,
\n$P=\\var{P}$,
\n$C=\\var{C}$,
\n$i=\\frac{\\var{ipa}\\%}{\\var{period[1]}}=\\frac{\\var{ipadec}}{\\var{period[1]}}$,
\nand therefore we have
\n$\\displaystyle \\var{P}=\\frac{\\var{C}}{\\left(\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)}\\left(1-\\frac{1}{\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)^n}\\right)\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)$
\nwhich we need to rearrange to solve for $n$. Note there are various methods that could be used, the following is one approach that uses the most basic operations and logs. Alternatively, you could use a faster, more sophisticated approach using negative powers or reciprocals.
\n\nWe want to get $n$ by itself. We start by multiplying both sides by $\\left(\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)$ (to remove the division by it on the right hand side)
\n$\\displaystyle \\var{P}\\left(\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)=\\var{C}\\left(1-\\frac{1}{\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)^n}\\right)\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)$
\nNow we will divide both sides by $\\var{C}$ (to remove the multiplication by it on the right hand side)
\n$\\displaystyle \\frac{\\var{P}}{\\var{C}}\\left(\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)=\\left(1-\\frac{1}{\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)^n}\\right)\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)$
\nNow we will divide both sides by $\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)$ (to remove the multiplication by it on the right hand side)
\n$\\displaystyle \\frac{\\frac{\\var{P}}{\\var{C}}\\left(\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)}{\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)}=\\left(1-\\frac{1}{\\left(1+\\simplify[unitDenominator]{{ipadec}/{period[1]}}\\right)^n}\\right)$
\nTidying this up a bit (including using a calculator) we have
\n\n$\\displaystyle \\frac{\\var{nnum}}{\\var{nden}}=1-\\frac{1}{\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)^n}$
\nWe add $\\displaystyle\\frac{1}{\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)^n}$ to both sides
\n$\\displaystyle \\frac{\\var{nnum}}{\\var{nden}}+\\frac{1}{\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)^n}=1$
\nThen subtract $\\displaystyle \\frac{\\var{nnum}}{\\var{nden}}$ from both sides
\n$\\displaystyle \\frac{1}{\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)^n}=1-\\frac{\\var{nnum}}{\\var{nden}}$
\nTidying this up we get
\n$\\displaystyle \\frac{1}{\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)^n}=\\frac{\\var{compnum}}{\\var{nden}}$
\nWe multiply both sides by $\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)^n$ so that the $n$ is not on the bottom of the fraction
\n$\\displaystyle 1=\\frac{\\var{compnum}}{\\var{nden}}{\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)^n}$
\nNow multiply both sides by $\\var{nden}$ to remove the division by it
\n$\\displaystyle \\var{nden}=\\var{compnum}\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)^n$
\nNow divide both sides by $\\var{compnum}$ to remove the multiplication by it
\n$\\displaystyle \\frac{\\var{nden}}{\\var{compnum}}=\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)^n$
\nNow we take the log of both sides
\n$\\displaystyle \\log\\left(\\frac{\\var{nden}}{\\var{compnum}}\\right)=\\log\\left(\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)^n\\right)$
\nIn order to apply the log law that allows us to bring the power, $n$, down (that is, $\\log(x^n)=n\\log(x)$)
\n$\\displaystyle \\log\\left(\\frac{\\var{nden}}{\\var{compnum}}\\right)=n\\log\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)$
\nFinally, we divide both sides by $\\log\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)$ to get $n$ by itself
\n$\\displaystyle \\frac{\\log\\left(\\frac{\\var{nden}}{\\var{compnum}}\\right)}{\\log\\left(\\frac{\\var{nRnum}}{\\var{nRden}}\\right)}=n$
\n\nCalculating this we find
\n$n\\approx \\var{n}$
\nbut we don't round this in the traditional sense since if we rounded down we wouldn't have paid off the entire loan, therefore we always round up to the nearest whole number.
\nThat is, we need to wait $\\var{nrounded}$ time periods.
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\n(log(1/(1-P*(ipadec/period[1])/C)))/(log((1+ipadec/period[1]))) gives complex after applying ceiling?!
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