// Numbas version: finer_feedback_settings {"name": "David's copy of Solving exponential equations using logs", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variablesTest": {"condition": "", "maxRuns": 100}, "functions": {}, "name": "David's copy of Solving exponential equations using logs", "tags": ["exp", "exponential", "exponentials", "logarithm", "logarithms", "logs", "solving", "Solving equations", "solving equations"], "parts": [{"gaps": [{"checkingaccuracy": 0.001, "checkvariablenames": false, "variableReplacements": [], "marks": 1, "vsetrangepoints": 5, "showpreview": true, "expectedvariablenames": [], "answer": "{d}*log({frac})/log({b})", "checkingtype": "absdiff", "scripts": {}, "showFeedbackIcon": true, "vsetrange": [0, 1], "type": "jme", "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst"}], "scripts": {}, "steps": [{"scripts": {}, "prompt": "
We start solving the equation one operation at a time by doing the inverse to both sides, when we get to undoing the exponential we apply a log to both sides, we then use a log law and continue solving.
\n\n$\\var{a}$ | \n$=$ | \n$\\simplify{{p}({b})^(n/{d})+{c}}$ | \n\n |
$\\simplify{{a-c}}$ | \n$=$ | \n$\\simplify{{p}({b})^(n/{d})}$ | \n(subtract $\\var{c}$ from both sides) | \n
$\\var{frac}$ | \n$=$ | \n$\\simplify{{b}^(n/{d})}$ | \n(divide both sides by $\\var{p}$) | \n
$\\log(\\var{frac})$ | \n$=$ | \n$\\log(\\var{b}^{\\frac{n}{\\var{d}}})$ | \n(take the log of both sides) | \n
\n | $=$ | \n$\\frac{n}{\\var{d}}\\log(\\var{b})$ | \n(use a log law) | \n
\n | \n | \n | \n |
$\\displaystyle{\\frac{\\log(\\var{frac})}{\\log(\\var{b})}}$ | \n$=$ | \n$\\frac{n}{\\var{d}}$ | \n(divide both sides by $\\log(\\var{b})$) | \n
\n | \n | \n | \n |
$\\displaystyle{\\frac{\\var{d}\\log(\\var{frac})}{\\log(\\var{b})}}$ | \n$=$ | \n$n$ | \n(multiply both sides by $\\var{d}$) | \n
Solve the following equation for $n$
\n$\\displaystyle{\\simplify{{a}={p}({b})^(n/{d})+{c}}}.$
\n\n$n=$ [[0]]
\n\n
Note: Typing $\\log(5)$ will input the value $\\log_{10}(5)$, whereas $\\log5$ will not work.
Note: Typing $\\ln(5)$ will input the value $\\log_e(5)$, whereas $\\ln5$ will not work.
We start solving the equation one operation at a time by doing the inverse to both sides, when we get to undoing the exponential we apply a log to both sides, we then use a log law and continue solving.
\n\n${\\var{FV}}$ | \n$=$ | \n$\\displaystyle{\\simplify{{pay}((1+{int})^n-1)/{int}}}$ | \n\n |
$\\simplify{{FV*int}}$ | \n$=$ | \n$\\displaystyle{\\simplify{{pay}((1+{int})^n-1)}}$ | \n(multiply both sides by $\\var{int}$) | \n
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int}/{pay}}}$ | \n$=$ | \n$\\displaystyle{\\simplify{(1+{int})^n-1}}$ | \n(divide both sides by $\\var{pay}$) | \n
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int}/{pay}+1}}$ | \n$=$ | \n$\\displaystyle{\\simplify{(1+{int})^n}}$ | \n(add $1$ to both sides) | \n
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}}$ | \n$=$ | \n$\\displaystyle{\\simplify{(1+{int})^n}}$ | \n(tidy up left hand side) | \n
\n | \n | \n | \n |
$\\displaystyle{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}$ | \n$=$ | \n$\\displaystyle{\\log\\left((\\var{1+int})^n\\right)}$ | \n(take the log of both sides) | \n
\n | \n | \n | \n |
$\\displaystyle{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}$ | \n$=$ | \n$\\displaystyle{n\\log(\\var{1+int})}$ | \n(use a log law) | \n
\n | \n | \n | \n |
$\\displaystyle{\\frac{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}{\\log(\\var{1+int})}}$ | \n$=$ | \n$n$ | \n(divide both sides by $\\log(\\var{1+int})$) | \n
Solve the following equation for $n$
\n$\\displaystyle{\\simplify{{FV}={pay}((1+{int})^n-1)/{int}}}.$
\n\n$n=$ [[0]]
\n\n
Note: Typing $\\log(5)$ will input the value $\\log_{10}(5)$, whereas $\\log5$ will not work.
Note: Typing $\\ln(5)$ will input the value $\\log_e(5)$, whereas $\\ln5$ will not work.