// Numbas version: exam_results_page_options
{"name": "Joshua's copy of Logarithms: inverses", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"name": "Joshua's copy of Logarithms: inverses", "advice": "", "statement": "<p>Simplify the following expressions:</p>", "metadata": {"description": "<p>Simplifying expressions&nbsp;such as $b^{\\log_b(x)}$ and $b^{\\log_b(x)}$.</p>", "licence": "Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International"}, "variablesTest": {"maxRuns": 100, "condition": ""}, "tags": [], "ungrouped_variables": ["b", "d", "c", "a", "list", "p", "q", "r"], "variables": {"q": {"definition": "list[1]", "templateType": "anything", "name": "q", "description": "<p>q</p>", "group": "Ungrouped variables"}, "c": {"definition": "random(2,3,4,5,6,7,-2,-3,-4,-5,1/2,1/5,1/4,-1/2,-1/4,-1/5)", "templateType": "anything", "name": "c", "description": "<p>c</p>", "group": "Ungrouped variables"}, "a": {"definition": "random(2..7)", "templateType": "anything", "name": "a", "description": "<p>a</p>", "group": "Ungrouped variables"}, "list": {"definition": "shuffle([0,0,random(-12..12 except 0),random(-12..12 except 0),random(-12..12 except 0)])", "templateType": "anything", "name": "list", "description": "", "group": "Ungrouped variables"}, "r": {"definition": "list[2]", "templateType": "anything", "name": "r", "description": "", "group": "Ungrouped variables"}, "b": {"definition": "random(2, 3, 5, 7, 11, e)", "templateType": "anything", "name": "b", "description": "", "group": "Ungrouped variables"}, "d": {"definition": "random(-12..12 except b)", "templateType": "anything", "name": "d", "description": "", "group": "Ungrouped variables"}, "p": {"definition": "list[0]", "templateType": "anything", "name": "p", "description": "<p>p</p>", "group": "Ungrouped variables"}}, "rulesets": {}, "functions": {}, "parts": [{"gaps": [{"variableReplacementStrategy": "originalfirst", "scripts": {}, "showFeedbackIcon": true, "variableReplacements": [], "type": "jme", "checkingaccuracy": 0.001, "vsetrange": [0, 1], "expectedvariablenames": [], "showpreview": true, "notallowed": {"showStrings": false, "partialCredit": 0, "strings": ["log", "ln", "e"], "message": "<p>This expression can be simplified to not contain&nbsp;logs or exponentials.</p>"}, "answer": "x+{d}", "showCorrectAnswer": true, "checkvariablenames": false, "vsetrangepoints": 5, "checkingtype": "absdiff", "marks": 1}], "variableReplacementStrategy": "originalfirst", "scripts": {}, "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "gapfill", "variableReplacements": [], "marks": 0, "stepsPenalty": "1", "steps": [{"variableReplacementStrategy": "originalfirst", "scripts": {}, "showFeedbackIcon": true, "variableReplacements": [], "type": "information", "showCorrectAnswer": true, "marks": 0, "prompt": "<p>You might recall that&nbsp;$b^{\\log_{b}(a)}=a$ (for $a&gt;0$), and therefore</p>\n<p>\\[\\var{b}^{\\log_{\\var{b}}(\\simplify{x+{d}})}=\\simplify{x+{d}}, \\quad\\text{ for }x&gt;\\var{-d}.\\]</p>\n<p>We can think of the logarithm and the exponential (of the same base)&nbsp;<em>cancelling</em>&nbsp;each other out, or&nbsp;<em>undoing</em>&nbsp;each other since they are the&nbsp;<em>inverse</em>&nbsp;of each other. Some people might prefer to see more evidence of this, so here is a longer explanation:</p>\n<p>Let $\\simplify{log(x+{d},{b})}=n$.&nbsp;Recall the definition of log says that this is equivalent to \\[\\simplify{{b}^n=x+{d}}.\\] Now substitute&nbsp;$\\simplify{log(x+{d},{b})}$ for&nbsp;$n$ into this equation (since we defined them to be equal) and we have&nbsp;</p>\n<p>\\[\\simplify{{b}^(log(x+{d},{b}))=x+{d}}\\]</p>\n<p></p>"}], "prompt": "<p>$\\var{b}^{\\log_{\\var{b}}(\\simplify{x+{d}})}=$&nbsp;[[0]], for $x&gt;\\var{-d}$.</p>"}, {"gaps": [{"variableReplacementStrategy": "originalfirst", "vsetrange": [0, 1], "showFeedbackIcon": true, "variableReplacements": [], "answersimplification": "fractionNumbers", "type": "jme", "checkingaccuracy": 0.001, "scripts": {}, "showpreview": true, "notallowed": {"showStrings": false, "partialCredit": 0, "strings": ["log", "ln", "e"], "message": "<p>This expression can be simplified to not contain&nbsp;logs or exponentials.</p>"}, "answer": "(x^{a}+{d})^{c}", "showCorrectAnswer": true, "checkvariablenames": false, "vsetrangepoints": 5, "checkingtype": "absdiff", "expectedvariablenames": [], "marks": 1}], "variableReplacementStrategy": "originalfirst", "scripts": {}, "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "gapfill", "variableReplacements": [], "marks": 0, "stepsPenalty": "1", "steps": [{"variableReplacementStrategy": "originalfirst", "scripts": {}, "showFeedbackIcon": true, "variableReplacements": [], "type": "information", "showCorrectAnswer": true, "marks": 0, "prompt": "<p>Recall that $\\ln$ denotes&nbsp;$\\log_e$.</p>\n<p>Here is one way to&nbsp;do&nbsp;this simplification (use the log laws $b\\log(a)=\\log(a^b)$ and $b^{\\log_b(a)}=a$):</p>\n<p>$\\begin{align}e^{\\var[fractionNumbers]{c}\\ln(\\simplify{x^{a}+{d}})}&amp;=e^{\\ln\\left(\\simplify[fractionNumbers]{(x^{a}+{d})^{c}}\\right)}\\\\&amp;=\\simplify[fractionNumbers]{(x^{a}+{d})^{c}}\\end{align}$</p>\n<p>Another way (use the index law $a^{bc}=(a^b)^c$ and&nbsp;$b^{\\log_b(a)}=a$):</p>\n<p>$\\begin{align}e^{\\var[fractionNumbers]{c}\\ln(\\simplify{x^{a}+{d}})}&amp;=\\left(e^{\\ln(\\simplify{x^{a}+{d}})}\\right)^\\var[fractionNumbers]{c}\\\\&amp;=\\simplify[fractionNumbers]{(x^{a}+{d})^{c}}\\end{align}$</p>\n<p></p>"}], "prompt": "<p>$e^{\\var[fractionNumbers]{c}\\ln(\\simplify{x^{a}+{d}})}=$&nbsp;[[0]], for $\\simplify{x^{a}+{d}}&gt;0$.</p>"}, {"gaps": [{"variableReplacementStrategy": "originalfirst", "scripts": {}, "showFeedbackIcon": true, "variableReplacements": [], "type": "jme", "checkingaccuracy": 0.001, "vsetrange": [0, 1], "expectedvariablenames": [], "showpreview": true, "notallowed": {"showStrings": false, "partialCredit": 0, "strings": ["ln", "log", "e"], "message": "<p>This expression can be simplified to not contain&nbsp;logs or exponentials.</p>"}, "answer": "{p}x^2+{q}x+{r}", "showCorrectAnswer": true, "checkvariablenames": false, "vsetrangepoints": 5, "checkingtype": "absdiff", "marks": 1}], "variableReplacementStrategy": "originalfirst", "scripts": {}, "showCorrectAnswer": true, "showFeedbackIcon": true, "type": "gapfill", "variableReplacements": [], "marks": 0, "stepsPenalty": "1", "steps": [{"variableReplacementStrategy": "originalfirst", "scripts": {}, "showFeedbackIcon": true, "variableReplacements": [], "type": "information", "showCorrectAnswer": true, "marks": 0, "prompt": "<p>Recall that $\\ln$ denotes&nbsp;$\\log_e$.</p>\n<p>To simplify $\\displaystyle \\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)$ we might think of&nbsp;the logarithm and the exponential (of the same base)&nbsp;<em>cancelling</em>&nbsp;each other out, or&nbsp;<em>undoing</em>&nbsp;each other since they are the&nbsp;<em>inverse</em>&nbsp;of each other. Some people might prefer to see more evidence of this, so here is a longer explanation:</p>\n<p>Let $\\displaystyle \\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)=n$. Recall the definition of log says that this is equivalent to \\[e^n=e^\\simplify{{p}x^2+{q}x+{r}}\\]</p>\n<p>Therefore $n=\\simplify{{p}x^2+{q}x+{r}}$ and so we can conclude</p>\n<p>\\[\\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)=\\simplify{{p}x^2+{q}x+{r}}.\\]</p>"}], "prompt": "<p>$\\displaystyle \\ln\\left(e^{\\simplify{{p}x^2+{q}x+{r}}}\\right)=$&nbsp;[[0]]</p>"}], "preamble": {"js": "", "css": ""}, "variable_groups": [], "extensions": [], "type": "question", "contributors": [{"name": "Joshua Capel", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1479/"}]}]}], "contributors": [{"name": "Joshua Capel", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1479/"}]}