// Numbas version: exam_results_page_options {"name": "Regina's copy of Logarithms: Solving Equations 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false}, "question_groups": [{"pickingStrategy": "all-ordered", "questions": [{"ungrouped_variables": ["a", "c", "b", "d", "s", "sol2", "sol1"], "variablesTest": {"condition": "", "maxRuns": 100}, "extensions": [], "functions": {}, "preamble": {"css": "", "js": ""}, "tags": [], "variable_groups": [], "parts": [{"gaps": [{"vsetrange": [0, 1], "variableReplacementStrategy": "originalfirst", "showFeedbackIcon": true, "notallowed": {"strings": ["."], "partialCredit": 0, "message": "

Input as an integer, not as a decimal.

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Three rules for logs should be used:

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1. $\\log_a(x^q)=q\\log_a(x)$

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2. $\\log_a(\\frac{x}{y})=\\log_a(x)-\\log_a(y)$

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3. $a^x=y \\iff \\log_a y=x$

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So use rule 1 followed by rules 2 and 3 to get an equation for $x$.

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\$2\\log_{\\var{a}}(\\simplify{x+{b}})- \\log_{\\var{a}}(\\simplify{(x+{c})})=\\var{d}\$

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

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If you want help in solving the equation, click on Show steps. If you do so then you will lose 1 mark.

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Input all numbers as fractions or integers and not as decimals.

We use the following rules for logs:

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1. $\\log_a(x^q)=q\\log_a(x)$

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2. $\\log_a(\\frac{x}{y})=\\log_a(x)-\\log_a(y)$

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3. $a^x=y \\iff \\log_a y=x$

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Using rule 1 we get
\$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})})\$
Using rule 2 gives
\$\\log_{\\var{a}}(\\simplify{(x+{b})^2})- \\log_{\\var{a}}(\\simplify{(x+{c})})=\\log_{\\var{a}}\\left(\\simplify{(x+{b})^2/(x+{c})}\\right)\$
So the equation to solve becomes:
\$\\log_{\\var{a}}\\left(\\simplify{(x+{b})^2/(x+{c})}\\right)=\\var{d}\$
and using rule 3 this gives:
\$\\begin{eqnarray} \\simplify{(x+{b})^2/(x+{c})}&=&{\\var{a}}^{\\var{d}}\\Rightarrow\\\\ \\simplify{(x+{b})^2}&=&{\\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}}&=&0 \\end{eqnarray} \$
Solving this quadratic we get two solutions:

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$x=\\var{sol1}$ and $x=\\var{sol2}$

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We should check that these solutions gives positive values for $\\simplify{x+{b}}$ and $\\simplify{x+{c}}$ as otherwise the logs are not defined.

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The value $x=\\var{sol1}$ gives:

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Substituting this value for $x$ into $\\log_{\\var{a}}(\\simplify{x+{b}})$ we get $\\log_{\\var{a}}(\\simplify{{2*a^d}})$ so OK.

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Substituting this value for $x$ into $\\log_{\\var{a}}(\\simplify{x+{c}})$ we get $\\log_{\\var{a}}(\\simplify{{4*a^d}})$ so OK.

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Hence $x=\\var{sol1}$ is a solution to our original equation.

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The value $x=\\var{sol2}$ gives:

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Substituting this value for $x$ into $\\log_{\\var{a}}(\\simplify{x+{b}})$ we get $\\log_{\\var{a}}(\\simplify{{-a^d}})$ so NOT OK.

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Substituting this value for $x$ into $\\log_{\\var{a}}(\\simplify{x+{c}})$ we get $\\log_{\\var{a}}(\\simplify{{a^d}})$ so OK.

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Hence $x=\\var{sol2}$ is NOT a solution to our original equation as $\\log_{\\var{a}}(\\simplify{x+{b}})$ is not defined for this value of $x$.

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So there is only one solution $x=\\var{sol1}$.

Solve for $x$: $\\displaystyle 2\\log_{a}(x+b)- \\log_{a}(x+c)=d$.

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Make sure that your choice is a solution by substituting back into the equation.

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Solve the following equation for $x$.

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Input your answer as a fraction or an integer as appropriate and not as a decimal.

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