// Numbas version: exam_results_page_options
{"name": "Simon's copy of Logarithms: Solving equations 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"variable_groups": [], "name": "Simon's copy of Logarithms: Solving equations 2", "tags": [], "variablesTest": {"condition": "", "maxRuns": 100}, "functions": {}, "extensions": [], "preamble": {"css": "", "js": ""}, "advice": "<p>We use the following rules for logs:</p>\n<p>1.&nbsp;$\\log_a(x^q)=q\\log_a(x)$</p>\n<p>2. $\\log_a(\\frac{x}{y})=\\log_a(x)-\\log_a(y)$</p>\n<p>3.&nbsp;$a^x=y \\iff \\log_a y=x$</p>\n<p>Using rule 1 we get<br/>\\[2\\log_{\\var{a}}(\\simplify{x+{b}})- \\log_{\\var{a}}(\\simplify{(x+{c})})=\\log_{\\var{a}}((\\simplify{x+{b}})^2)- \\log_{\\var{a}}(\\simplify{(x+{c})})\\]<br/>Using rule 2 gives<br/>\\[\\log_{\\var{a}}(\\simplify{(x+{b})^2})- \\log_{\\var{a}}(\\simplify{(x+{c})})=\\log_{\\var{a}}\\left(\\simplify{(x+{b})^2/(x+{c})}\\right)\\]<br/>So the equation to solve becomes:<br/>\\[\\log_{\\var{a}}\\left(\\simplify{(x+{b})^2/(x+{c})}\\right)=\\var{d}\\]<br/>and using rule 3 this gives:<br/>\\[ \\begin{eqnarray} \\simplify{(x+{b})^2/(x+{c})}&amp;=&amp;{\\var{a}}^{\\var{d}}\\Rightarrow\\\\ \\simplify{(x+{b})^2}&amp;=&amp;{\\var{a}}^{\\var{d}}(\\simplify{x+{c}})=\\simplify{{a^d}(x+{c})}\\Rightarrow\\\\ \\simplify{x^2+{2*b-a^(d)}x+{b^2-a^(d)*c}}&amp;=&amp;0 \\end{eqnarray} \\]<br/>Solving this quadratic we get two solutions:</p>\n<p>$x=\\var{sol1}$ and $x=\\var{sol2}$</p>\n<p>We should check that these solutions gives positive values for $\\simplify{x+{b}}$ and $\\simplify{x+{c}}$ as otherwise the logs are not defined.</p>\n<p><strong>The value $x=\\var{sol1}$ gives:&nbsp;</strong></p>\n<p>Substituting this value for $x$ into $\\log_{\\var{a}}(\\simplify{x+{b}})$ we get $\\log_{\\var{a}}(\\simplify{{2*a^d}})$ which is OK.</p>\n<p>Substituting this value for $x$ into $\\log_{\\var{a}}(\\simplify{x+{c}})$ we get $\\log_{\\var{a}}(\\simplify{{4*a^d}})$&nbsp;which is OK.</p>\n<p>Hence $x=\\var{sol1}$ is a solution to our original equation.</p>\n<p><strong>The value $x=\\var{sol2}$ gives:</strong></p>\n<p>Substituting this value for $x$ into $\\log_{\\var{a}}(\\simplify{x+{b}})$ we get $\\log_{\\var{a}}(\\simplify{{-a^d}})$ which is&nbsp;<span class=\"caps\">NOT</span> OK (we cannot take logs of a <strong>negative</strong> number)</p>\n<p>Substituting this value for $x$ into $\\log_{\\var{a}}(\\simplify{x+{c}})$ we get $\\log_{\\var{a}}(\\simplify{{a^d}})$&nbsp;which is OK.</p>\n<p>Hence $x=\\var{sol2}$ is <span class=\"caps\">NOT</span> a solution to our original equation as $\\log_{\\var{a}}(\\simplify{x+{b}})$ is not defined for this value of $x$.</p>\n<p>So there is only one solution $x=\\var{sol1}$.</p>", "variables": {"sol2": {"group": "Ungrouped variables", "name": "sol2", "description": "", "templateType": "anything", "definition": "-c+a^d"}, "sol1": {"group": "Ungrouped variables", "name": "sol1", "description": "", "templateType": "anything", "definition": "c-2*b"}, "c": {"group": "Ungrouped variables", "name": "c", "description": "", "templateType": "anything", "definition": "b+2*a^(d)"}, "d": {"group": "Ungrouped variables", "name": "d", "description": "", "templateType": "anything", "definition": "random(1,2)"}, "b": {"group": "Ungrouped variables", "name": "b", "description": "", "templateType": "anything", "definition": "s*random(1..20)"}, "s": {"group": "Ungrouped variables", "name": "s", "description": "", "templateType": "anything", "definition": "random(1,-1)"}, "a": {"group": "Ungrouped variables", "name": "a", "description": "", "templateType": "anything", "definition": "random(2,3)"}}, "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "statement": "\n  <p>Solve the following equation for $x$.</p>\n  <p>Input your answer as a fraction or an integer as appropriate and not&nbsp;as a decimal.</p>\n  ", "parts": [{"sortAnswers": false, "scripts": {}, "gaps": [{"answer": "{sol1}", "vsetRangePoints": 5, "scripts": {}, "checkingType": "absdiff", "showPreview": true, "customMarkingAlgorithm": "", "answerSimplification": "std", "notallowed": {"showStrings": false, "strings": ["."], "partialCredit": 0, "message": "<p>Input as an integer, not as a decimal.</p>"}, "unitTests": [], "showCorrectAnswer": true, "expectedVariableNames": [], "vsetRange": [0, 1], "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "type": "jme", "checkVariableNames": false, "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": "3", "checkingAccuracy": 0.0001, "failureRate": 1}], "stepsPenalty": 1, "customMarkingAlgorithm": "", "steps": [{"prompt": "<p>Three rules for logs should be used:</p>\n<p>1.&nbsp;$\\log_a(x^q)=q\\log_a(x)$</p>\n<p>2. $\\log_a(\\frac{x}{y})=\\log_a(x)-\\log_a(y)$</p>\n<p>3.&nbsp;$a^x=y \\iff \\log_a y=x$</p>\n<p>So use rule 1 followed by rules 2 and 3 to get an equation for $x$.</p>", "extendBaseMarkingAlgorithm": true, "scripts": {}, "showFeedbackIcon": true, "variableReplacementStrategy": "originalfirst", "customMarkingAlgorithm": "", "type": "information", "variableReplacements": [], "marks": 0, "unitTests": [], "showCorrectAnswer": true}], "unitTests": [], "showCorrectAnswer": true, "prompt": "\n      <p>\\[2\\log_{\\var{a}}(\\simplify{x+{b}})- \\log_{\\var{a}}(\\simplify{(x+{c})})=\\var{d}\\]</p>\n      <p>$x=\\;$ [[0]].</p>\n      <p>If you want help in solving the equation, click on Show steps. If you do so then you will lose 1 mark.</p>\n      <p>Input all numbers as fractions or integers and not as decimals.</p>\n      ", "extendBaseMarkingAlgorithm": true, "showFeedbackIcon": true, "type": "gapfill", "variableReplacementStrategy": "originalfirst", "variableReplacements": [], "marks": 0}], "ungrouped_variables": ["a", "c", "b", "d", "s", "sol2", "sol1"], "metadata": {"licence": "Creative Commons Attribution 4.0 International", "description": "\n            \t\t<p>Solve for $x$: $\\displaystyle 2\\log_{a}(x+b)- \\log_{a}(x+c)=d$.&nbsp;</p>\n            \t\t        <p>Make sure that your choice is a solution by substituting back into the equation.</p>\n            \t\t"}, "type": "question", "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}]}], "contributors": [{"name": "Julie Crowley", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/113/"}, {"name": "Simon Thomas", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/3148/"}]}