// Numbas version: finer_feedback_settings {"name": "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": [{"variable_groups": [{"variables": ["FV", "pay", "int", "periods", "test"], "name": "b"}], "preamble": {"js": "", "css": ""}, "ungrouped_variables": ["a", "p", "c", "b", "frac", "logfrac", "logb", "d", "amc", "n"], "functions": {}, "extensions": [], "variablesTest": {"condition": "", "maxRuns": 100}, "metadata": {"description": "", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "statement": "", "variables": {"int": {"templateType": "anything", "description": "", "definition": "random(0.01..0.10#0.01)", "group": "b", "name": "int"}, "n": {"templateType": "anything", "description": "", "definition": "d*logfrac/logb", "group": "Ungrouped variables", "name": "n"}, "test": {"templateType": "anything", "description": "", "definition": "fv*int/pay", "group": "b", "name": "test"}, "amc": {"templateType": "anything", "description": "", "definition": "a-c", "group": "Ungrouped variables", "name": "amc"}, "p": {"templateType": "anything", "description": "", "definition": "random(2,5,10,20,(a-c)/2,(a-c)/5,(a-c)/10,(a-c)/20)", "group": "Ungrouped variables", "name": "p"}, "FV": {"templateType": "anything", "description": "", "definition": "random(20000..100000#500)", "group": "b", "name": "FV"}, "c": {"templateType": "anything", "description": "", "definition": "random(20..a-10#20)", "group": "Ungrouped variables", "name": "c"}, "frac": {"templateType": "anything", "description": "", "definition": "(a-c)/p", "group": "Ungrouped variables", "name": "frac"}, "b": {"templateType": "anything", "description": "", "definition": "random(1.05..1.5#0.05)", "group": "Ungrouped variables", "name": "b"}, "a": {"templateType": "anything", "description": "", "definition": "random(1000..2000#20)", "group": "Ungrouped variables", "name": "a"}, "d": {"templateType": "anything", "description": "", "definition": "random(2,4,3,12,26,52)", "group": "Ungrouped variables", "name": "d"}, "periods": {"templateType": "anything", "description": "", "definition": "log(FV*int/pay+1)/log(1+int)", "group": "b", "name": "periods"}, "logfrac": {"templateType": "anything", "description": "", "definition": "log(frac)", "group": "Ungrouped variables", "name": "logfrac"}, "pay": {"templateType": "anything", "description": "", "definition": "random(100..2000#100)", "group": "b", "name": "pay"}, "logb": {"templateType": "anything", "description": "", "definition": "log(b)", "group": "Ungrouped variables", "name": "logb"}}, "tags": [], "parts": [{"customName": "", "customMarkingAlgorithm": "", "sortAnswers": false, "scripts": {}, "prompt": "
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$\\var{a}$ | \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$\\displaystyle{\\simplify{{FV}={pay}((1+{int})^n-1)/{int}}}.$
\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.
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