// 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.

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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
$\\var{a}$$=$$\\simplify{{p}({b})^(n/{d})+{c}}$
$\\simplify{{a-c}}$$=$$\\simplify{{p}({b})^(n/{d})}$(subtract $\\var{c}$ from both sides) 
$\\var{frac}$$=$$\\simplify{{b}^(n/{d})}$(divide both sides by $\\var{p}$)
$\\log(\\var{frac})$$=$$\\log(\\var{b}^{\\frac{n}{\\var{d}}})$(take the log of both sides)
$=$ $\\frac{n}{\\var{d}}\\log(\\var{b})$(use a log law)
$\\displaystyle{\\frac{\\log(\\var{frac})}{\\log(\\var{b})}}$$=$$\\frac{n}{\\var{d}}$(divide both sides by $\\log(\\var{b})$)
$\\displaystyle{\\frac{\\var{d}\\log(\\var{frac})}{\\log(\\var{b})}}$$=$$n$(multiply both sides by $\\var{d}$)
\n

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Solve the following equation for $n$

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$\\displaystyle{\\simplify{{a}={p}({b})^(n/{d})+{c}}}.$

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$n=$ [[0]]

\n

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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.

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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\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n\n
${\\var{FV}}$$=$$\\displaystyle{\\simplify{{pay}((1+{int})^n-1)/{int}}}$
$\\simplify{{FV*int}}$$=$$\\displaystyle{\\simplify{{pay}((1+{int})^n-1)}}$(multiply both sides by $\\var{int}$) 
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int}/{pay}}}$$=$$\\displaystyle{\\simplify{(1+{int})^n-1}}$(divide both sides by $\\var{pay}$)
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int}/{pay}+1}}$$=$$\\displaystyle{\\simplify{(1+{int})^n}}$(add $1$ to both sides)
$\\displaystyle{\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}}$$=$ $\\displaystyle{\\simplify{(1+{int})^n}}$(tidy up left hand side)
$\\displaystyle{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}$$=$$\\displaystyle{\\log\\left((\\var{1+int})^n\\right)}$(take the log of both sides)
$\\displaystyle{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}$$=$$\\displaystyle{n\\log(\\var{1+int})}$(use a log law)
$\\displaystyle{\\frac{\\log\\left(\\simplify[simplifyFractions]{{FV*int+pay}/{pay}}\\right)}{\\log(\\var{1+int})}}$$=$$n$(divide both sides by $\\log(\\var{1+int})$)
\n

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Solve the following equation for $n$

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$\\displaystyle{\\simplify{{FV}={pay}((1+{int})^n-1)/{int}}}.$

\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.

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