// Numbas version: finer_feedback_settings {"name": "Logarithms1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"functions": {}, "name": "Logarithms1", "tags": [], "advice": "\n

The rules for combining logs are

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\\[\\begin{eqnarray*}&1.&  \\log_b(ac)&=&\\log_b(a)+\\log_b(c)\\\\ \\\\ &2.&  \\log_b\\left(\\frac{a}{c}\\right)&=&\\log_b(a)-\\log_b(c)\\\\ \\\\ &3.&  \\log_b(a^r)&=&r\\log_b(a) \\end{eqnarray*} \\]

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We see that:

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\\[\\begin{eqnarray*}\\log_{\\var{b}}(\\var{a_1})-\\var{r_1}\\log_{\\var{b}}(\\var{a_2})+\\var{r_2}\\log_{\\var{b}}(\\var{a_3})&=&\\log_{\\var{b}}(\\var{a_1})-\\log_{\\var{b}}(\\var{a_2}^{\\var{r_1}})+\\log_{\\var{b}}(\\var{a_3}^{\\var{r_2}})\\mbox{ using 3.}\\\\&=&\\log_{\\var{b}}(\\var{a_1})-\\log_{\\var{b}}(\\var{a_2^r_1})+\\log_{\\var{b}}(\\var{a_3^r_2})\\\\&=&\\log_{\\var{b}}(\\var{a_1}\\times \\var{a_3^r_2})-\\log_{\\var{b}}(\\var{a_2^r_1}) \\mbox{ using 1.}\\\\&=&\\log_{\\var{b}}\\left(\\frac{\\var{a_1}\\times \\var{a_3^r_2}}{\\var{a_2^r_1}}\\right) \\mbox{ using 2.}\\\\&=&\\log_{\\var{b}}\\left(\\frac{\\var{a_1*a_3^r_2}}{\\var{a_2^r_1}}\\right)\\\\&=&\\log_{\\var{b}}\\left(\\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\\right)\\mbox{ on cancelling common factors}.\\end{eqnarray*}\\]

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Hence $\\displaystyle c=\\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}$.

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To calculate $\\displaystyle \\log_{\\var{b}}\\left(\\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\\right)$ to 4 decimal places we use the fact that for any positive base $b$:

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\\[\\log_b(c)=\\frac{\\ln(c)}{\\ln(b)}=\\frac{\\log_{10}(c)}{\\log_{10}(b)}\\]

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and we can use either of the log functions, $\\ln$ or $\\log_{10}$ on our calculators to find the value.

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Using $\\ln$ we find:

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\\[ \\log_{\\var{b}}\\left(\\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\\right)=\\frac{\\ln\\left(\\simplify[all,fractionnumbers]{{a_1*a_3^r_2}/{a_2^r_1}}\\right)}{\\ln(\\var{b})}=\\var{ans}\\] to 4 decimal places.

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Find a fraction or integer $c$ such that:

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$\\log_{\\var{b}}(\\var{a_1})-\\var{r_1}\\log_{\\var{b}}(\\var{a_2})+\\var{r_2}\\log_{\\var{b}}(\\var{a_3})=\\log_{\\var{b}}(c)$ 

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$c=\\;$[[0]].

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Input $c$ as an integer or as a fraction and not as a decimal.

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Now calculate $\\log_{\\var{b}}(c)$ to 4 decimal places:

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$\\log_{\\var{b}}(c)=\\;$[[1]].

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Answer the following question on logarithms.

", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"c": {"definition": "a_1*a_3^r_2/a_2^r_1", "name": "c"}, "b": {"definition": "random(2..9)", "name": "b"}, "a_3": {"definition": "random(2..8)", "name": "a_3"}, "a_2": {"definition": "random(2,4,8)*random(3,9)", "name": "a_2"}, "a_1": {"definition": "random(5..20)", "name": "a_1"}, "r_1": {"definition": "random(2..3)", "name": "r_1"}, "r_2": {"definition": "random(2..3)", "name": "r_2"}, "tol": {"definition": 0.0001, "name": "tol"}, "ans": {"definition": "precround(log(c)/log(b),4)", "name": "ans"}}, "metadata": {"notes": "\n \t\t

30/4/2013:

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Written for revision of logs in  a foundation course

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Given a sum of logs, all numbers are integers,

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$\\log_b(a_1)+\\alpha\\log_b(a_2)+\\beta\\log_b(a_3)$ write as $\\log_b(a)$ for some fraction $a$.

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Also calculate to 3 decimal places $\\log_b(a)$. 

\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Wan Mekwi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4058/"}]}]}], "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}, {"name": "Wan Mekwi", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/4058/"}]}