// Numbas version: finer_feedback_settings {"name": "Maths Support: Logarithms and solving equations", "navigation": {"onleave": {"action": "none", "message": ""}, "reverse": true, "allowregen": true, "preventleave": false, "browse": true, "showfrontpage": false, "showresultspage": "never"}, "duration": 0.0, "metadata": {"notes": "", "description": "

8 questions using logarithms. 7 questions use logarithms to solve equations.

", "licence": "Creative Commons Attribution 4.0 International"}, "timing": {"timeout": {"action": "none", "message": ""}, "timedwarning": {"action": "none", "message": ""}}, "shufflequestions": false, "questions": [], "percentpass": 0.0, "allQuestions": true, "pickQuestions": 0, "type": "exam", "feedback": {"showtotalmark": true, "advicethreshold": 0.0, "showanswerstate": true, "showactualmark": true, "allowrevealanswer": true, "enterreviewmodeimmediately": false, "showexpectedanswerswhen": "never", "showpartfeedbackmessageswhen": "always", "showactualmarkwhen": "always", "showtotalmarkwhen": "always", "showanswerstatewhen": "always", "showadvicewhen": "never"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": [{"name": "Hyperbolic Functions 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["cosh", "hyperbolic equations", "hyperbolic functions", "logarithms", "quadratic equation", "sinh", "solving equations", "solving hyperbolic equations", "solving quadratic equation"], "advice": "\n

Using \\[\\cosh(x)=\\frac{e^x+e^{-x}}{2},\\;\\;\\;\\sinh(x)=\\frac{e^x-e^{-x}}{2}\\] and substituting into the equation gives:

\\[\\simplify[std]{{a1+b1}e^x+{a1-b1}e^(-x)={2*c1}}\\]

On multiplying throughout by $e^x$, putting $y=e^x$ and tidying up the equation, we get:

\\[\\simplify[std]{{m1}* y ^ 2 + {m2} * y + {m3} = 0}\\]

This quadratic has solutions:

\\[y = \\simplify[std]{{d1}/{al1}},\\;\\;\\;y=\\simplify[std]{{d2}/{be1}}\\]

Since $y=e^x$ we see that the solutions in terms of $x$ are:
\\[x = \\ln\\left(\\simplify[std]{{d1}/{al1}}\\right)=\\var{tans2},\\;\\;\\;x=\\ln\\left(\\simplify[std]{{d2}/{be1}}\\right)=\\var{tans3}\\]
both to 3 decimal places.

Hence the minimum solution is $x=\\var{ans2}$ and the maximum solution is $x=\\var{ans3}$.

\n ", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\n \n \n

Input the solutions for $x$ here:(if the solutions are the same input the number twice)

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Least solution = [[0]]

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Greatest solution= [[1]]

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Input both to 3 decimal places.

\n \n \n ", "gaps": [{"minvalue": "ans2-tol", "type": "numberentry", "maxvalue": "ans2+tol", "marks": 1.0, "showPrecisionHint": false}, {"minvalue": "ans3", "type": "numberentry", "maxvalue": "ans3", "marks": 1.0, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}], "statement": "\n

Solve the following equation for $x$.

\\[\\simplify[std]{{a1}cosh(x)+{b1}sinh(x)={c1}}\\]

\n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"ans2": {"definition": "precround(ans12,3)", "name": "ans2"}, "ans3": {"definition": "precround(ans21,3)", "name": "ans3"}, "b1": {"definition": "s*(al1*be1-d1*d2)/2", "name": "b1"}, "d2": {"definition": "s6*random(1..8)", "name": "d2"}, "ans21": {"definition": "max(tans2,tans3)", "name": "ans21"}, "d1": {"definition": "if(al1*be1=d*d2,d+2,d)", "name": "d1"}, "al1": {"definition": "s5*random(2,4,6)", "name": "al1"}, "s6": {"definition": "random(1,-1)", "name": "s6"}, "s5": {"definition": 1.0, "name": "s5"}, "m1": {"definition": "round((a1+b1)/h)", "name": "m1"}, "m3": {"definition": "round((a1-b1)/h)", "name": "m3"}, "m2": {"definition": "round(-2*c1/h)", "name": "m2"}, "be1": {"definition": "s6*random(1..9)", "name": "be1"}, "tol": {"definition": 0.0, "name": "tol"}, "ans12": {"definition": "min(tans2,tans3)", "name": "ans12"}, "a2": {"definition": "(al1*be1+d1*d2)/2", "name": "a2"}, "c1": {"definition": "s*(d1*be1+d2*al1)/2", "name": "c1"}, "tans3": {"definition": "precround(ln(d2/be1),3)", "name": "tans3"}, "tans2": {"definition": "precround(ln(d1/al1),3)", "name": "tans2"}, "a1": {"definition": "abs(a2)", "name": "a1"}, "d": {"definition": "s5*random(2,4)", "name": "d"}, "h": {"definition": "gcf(gcf(abs(a1+b1),abs(a1-b1)),abs(2*c1))", "name": "h"}, "s": {"definition": "sign(a2)", "name": "s"}}, "metadata": {"notes": "\n \t\t

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Solve for $x$: $a\\cosh(x)+b\\sinh(x)=c$. There are two solutions for this example.

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Logarithms: Solving equations 1", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Steps", "algebra", "algebraic manipulation", "combining logarithms", "logarithm laws", "logarithms", "simplifying logarithms", "solving equations", "steps"], "advice": "\n

We use the following two rules for logs :

1. $\\log_a(b)-\\log_a(c)=\\log_a(b/c)$

2. $\\log_a(p)=r \\Rightarrow p=a^r$

Using rule 1 we get
\\[\\log_{\\var{a}}(x+\\var{b})- \\log_{\\var{a}}(\\simplify{(x+{c})})=\\log_{\\var{a}}\\left(\\simplify{(x+{b})/(x+{c})}\\right)\\]
So the equation to solve becomes:
\\[\\log_{\\var{a}}\\left(\\simplify{(x+{b})/(x+{c})}\\right)=\\var{d}\\]
and using rule 2 this gives:
\\[ \\begin{eqnarray} \\simplify{(x+{b})/(x+{c})}&=&{\\var{a}}^{\\var{d}}\\Rightarrow\\\\ x+\\var{b}&=&{\\var{a}}^{\\var{d}}(x+\\var{c})=\\simplify{{a^d}}(x+\\var{c})\\Rightarrow\\\\ \\simplify{{a^d-1}x}&=&\\simplify[std]{{b}-{c}*{a^d}={b-c*a^d}}\\Rightarrow\\\\ x&=&\\simplify{{b-c*a^d}/{a^d-1}} \\end{eqnarray} \\]
We should check that this solution gives positive values for $x+\\var{b}$ and $\\simplify{x+{c}}$ as otherwise the logs are not defined.

Substituting this value for $x$ into $\\log_{\\var{a}}(x+\\var{b})$ we get $\\log_{\\var{a}}(\\simplify{({b-c }{a^d})/{a^d-1}})$ so OK.

For $\\log_{\\var{a}}(\\simplify{x+{c}})$ we get on substituting for $x$, $\\log_{\\var{a}}(\\simplify{({b-c })/{a^d-1}})$ so OK.

Hence the value we found for $x$ is a solution to the original equation.

\n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers", "!noLeadingMinus"]}, "parts": [{"stepspenalty": 1.0, "prompt": "\n

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

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

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

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

\n ", "gaps": [{"notallowed": {"message": "

Input as a fraction or an integer, not as a decimal.

", "showstrings": false, "strings": ["."], "partialcredit": 0.0}, "checkingaccuracy": 0.0001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 2.0, "answer": "{b-c*a^d}/{a^d-1}", "type": "jme"}], "steps": [{"prompt": "\n

Two rules for logs should be used:

1. $\\log_a(b)-\\log_a(c)=\\log_a(b/c)$

2. $\\log_a(p)=r \\Rightarrow p=a^r$

Use rule 1 followed by rule 2 to get an equation for $x$.

\n ", "type": "information", "marks": 0.0}], "marks": 0.0, "type": "gapfill"}], "statement": "\n

Solve the following equation for $x$.

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

\n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2..5)", "name": "a"}, "c": {"definition": "b-random(1..20)", "name": "c"}, "b": {"definition": "random(1..20)", "name": "b"}, "d": {"definition": "random(1,2)", "name": "d"}}, "metadata": {"notes": "\n \t\t

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

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Logarithms: Solving equations 2", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["Steps", "algebra", "algebraic manipulation", "combining logarithms", "logarithm laws", "logarithms", "simplifying logarithms", "solving", "solving equations", "steps"], "advice": "\n

We use the following three rules for logs :

1. $n\\log_a(m)=\\log_a(m^n)$

2. $\\log_a(b)-\\log_a(c)=\\log_a(b/c)$

3. $\\log_a(p)=r \\Rightarrow p=a^r$

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:

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

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

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

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

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

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

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

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

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

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 $x$.

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

\n ", "rulesets": {"std": ["all", "!collectNumbers", "fractionNumbers"]}, "parts": [{"stepspenalty": 1.0, "prompt": "\n

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

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

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

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

\n ", "gaps": [{"notallowed": {"message": "

Input as an integer, not as a decimal.

Three rules for logs should be used:

1. $n\\log_a(m)=\\log_a(m^n)$

2. $\\log_a(b)-\\log_a(c)=\\log_a(b/c)$

3. $\\log_a(p)=r \\Rightarrow p=a^r$

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

\n ", "type": "information", "marks": 0.0}], "marks": 0.0, "type": "gapfill"}], "statement": "\n

Solve the following equation for $x$.

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

\n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(2,3)", "name": "a"}, "c": {"definition": "b+2*a^(d)", "name": "c"}, "b": {"definition": "s*random(1..20)", "name": "b"}, "d": {"definition": "random(1,2)", "name": "d"}, "s": {"definition": "random(1,-1)", "name": "s"}, "sol2": {"definition": "-c+a^d", "name": "sol2"}, "sol1": {"definition": "c-2*b", "name": "sol1"}}, "metadata": {"notes": "\n \t\t

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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.

\n \t\t", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Logarithms: Solving equations 3", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["exponential", "exponentiation", "laws of logarithms", "laws of logs", "log laws", "logarithm laws", "logarithm rules", "logarithms", "logs", "solving equations", "solving logarithmic equations"], "advice": "\n

First use one of the logarith laws which states (for logarithms to any base)

\\[\\log(a)-\\log(b)=\\log\\left(\\frac{a}{b}\\right)\\]

So the equation can be written as:

\\[\\log_{10}\\left(\\simplify[std]{({a1}x+{b1})/({c1}x+{d1})}\\right)=\\var{e1}\\]
Now exponentiate both sides to get:
\\[\\simplify[std,!otherNumbers]{({a1}x+{b1})/({c1}x+{d1})}=10^{\\var{e1}} \\Rightarrow \\simplify[std,!otherNumbers]{{a1}x+{b1}=10^{e1}({c1}x+{d1})}\\]
Collect together terms in $x$ on the left and everything else on the right of the equation gives:
\\[\\simplify[std,!otherNumbers]{x({a1}-10^{e1}*{c1})=10^{e1}*{d1}-{b1}}\\]
Finally rearrange to get:
\\[\\simplify[std]{x=(10^{e1}*{d1}-{b1})/({a1}-10^{e1}*{c1})={10^e1*d1-b1}/{a1-10^e1*c1}}\\]
which to 3 decimal places evaluates to
\\[x=\\var{ans}.\\]

\n ", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\n

Input the solution for $x$ here:

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

Input your answer to 3 decimal places.

\n ", "gaps": [{"minvalue": "ans-tol", "type": "numberentry", "maxvalue": "ans+tol", "marks": 1.0, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}], "statement": "\n

Solve the following equation for $x$.

\\[\\simplify[std]{log({a1}x+{b1})-log({c1}x+{d1})={e1}}\\]

\n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"e1": {"definition": "random(1..2)", "name": "e1"}, "a1": {"definition": "random(1..9)", "name": "a1"}, "b1": {"definition": "9+random(1..9)", "name": "b1"}, "tol": {"definition": 0.0, "name": "tol"}, "ans": {"definition": "precround(tans,3)", "name": "ans"}, "c1": {"definition": "d1*random(1..9)", "name": "c1"}, "tans": {"definition": "(d1*10^e1-b1)/(a1-c1*10^e1)", "name": "tans"}, "d1": {"definition": "random(1..9)", "name": "d1"}}, "metadata": {"notes": "\n \t\t

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Solve for $x$: $\\log(ax+b)-\\log(cx+d)=s$

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Logarithms: Solving equations 4", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["equations solved by using logarithms", "laws of logarithms", "logarithm laws", "logarithm rules", "logarithmic expressions", "logarithms", "solving equations by taking logarithms", "solving logarithmic equations"], "advice": "

Both parts of this question can be solved in a similar way, by taking logarithms of both sides of each equation.

Taking logs (to the base 10 in this case – but any base will do) of both sides of

\\[\\var{n}^{\\simplify{{a1}*x+{b1}}}=\\var{m}^{\\var{c1}x}\\]

gives on using the rule $\\log(a^b)=b\\log(a)$:

\\[\\begin{eqnarray*} \\simplify[std]{({a1}x+{b1})log({n})}&=&\\simplify[std]{{c1}*x*log({m})}\\\\ \\Rightarrow\\simplify[std]{x({a1}*log({n})-{c1}*log({m}))} &=&\\var{-b1}\\log(\\var{n})\\\\ \\Rightarrow x&=&\\simplify[std]{({-b1}*log({n}))/({a1}log({n})-{c1}*log({m}))}\\\\ &=& \\var{ans1}\\mbox{ to 3 decimal places} \\end{eqnarray*} \\]

Similarly, taking logs of both sides of:
\\[\\var{a2}^{\\var{b2}x^2}=\\var{c2}^{\\var{d2}x}\\]

gives:

\\[ \\simplify[std]{({b2}x^2)log({a2})}=\\simplify[std]{{d2}*x*log({c2})} \\Rightarrow \\simplify[std]{x({b2}*log({a2})x-{d2}*log({c2}))} =0\\]

and so the solutions are:

1. $x=0$

2. $\\displaystyle x=\\simplify[std,!fractionNumbers]{({d2}*log({c2}))/({b2}*log({a2})) = {ans2}}$ to 3 decimal places.

", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\n

\\[\\var{n}^{\\simplify{{a1}*x+{b1}}}=\\var{m}^{\\var{c1}x}\\]

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

Enter your answer to 3 decimal places.

\n ", "gaps": [{"minvalue": "{ans1}", "type": "numberentry", "maxvalue": "{ans1}", "marks": 1.0, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n

\\[\\var{a2}^{\\var{b2}x^2}=\\var{c2}^{\\var{d2}x}\\]

$x=\\;\\;$[[0]] $\\;\\;\\;$ (Enter the smallest value of $x$ here).

$x=\\;\\;$[[1]] $\\;\\;\\;$ (Enter the largest value of $x$ here).

Enter your answers to 3 decimal places.

\n ", "gaps": [{"minvalue": 0.0, "type": "numberentry", "maxvalue": 0.0, "marks": 0.5, "showPrecisionHint": false}, {"minvalue": "ans2", "type": "numberentry", "maxvalue": "ans2", "marks": 1.5, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}], "statement": "

Find all values of $x$ that satisfy the following equations:

", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"tc2": {"definition": "random(2..9)", "name": "tc2"}, "s3": {"definition": "random(1,-1)", "name": "s3"}, "ans1": {"definition": "precround(tans1,3)", "name": "ans1"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "m": {"definition": "if(tm=n,17,tm)", "name": "m"}, "d2": {"definition": "random(2..9)", "name": "d2"}, "n": {"definition": "random(2,3,5,7,11,13)", "name": "n"}, "a1": {"definition": "s1*random(1..9)", "name": "a1"}, "tm": {"definition": "random(2,3,5,7,11,13)", "name": "tm"}, "b1": {"definition": "random(1..9)", "name": "b1"}, "b2": {"definition": "random(2..9)", "name": "b2"}, "c2": {"definition": "if(tc2=a2,tc2+1,tc2)", "name": "c2"}, "c1": {"definition": "s3*random(2..9)", "name": "c1"}, "tans1": {"definition": "-b1*log(n)/(a1*log(n)-c1*log(m))", "name": "tans1"}, "ans2": {"definition": "precround(tans2,3)", "name": "ans2"}, "a2": {"definition": "random(2..9)", "name": "a2"}, "tans2": {"definition": "d2*log(c2)/(b2*log(a2))", "name": "tans2"}}, "metadata": {"notes": "\n \t\t

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Solve for $x$ each of the following equations: $n^{ax+b}=m^{cx}$ and $p^{rx^2}=q^{sx}$.

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Logarithms: Solving equations 5", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["exponential function", "logarithmic equation", "logarithms", "logs", "natural logarithms", "solving equations", "solving logarithmic equations"], "advice": "\n \n \n

We have on solving the equation $\\rho(t)=\\rho_0e^{kt}$ by taking natural logs of both sides that:
\\[k = \\frac{1}{t}\\ln\\left(\\frac{\\rho}{\\rho_0}\\right)\\]

\n \n \n \n

a)
\\[\\begin{eqnarray*}\n \n k&=& \\frac{1}{\\var{c}}\\ln\\left(\\frac{\\var{a*r} \\times 10^{\\var{b}}}{\\var{a} \\times 10^{\\var{b}}}\\right)\\\\\n \n &=& \\frac{1}{\\var{c}}\\ln(\\var{r})\\\\\n \n &=&\\var{kvalue}\n \n \\end{eqnarray*}\n \n \\] to 3 decimal places.

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b)
\\[\\begin{eqnarray*}\n \n k&=& \\frac{1}{\\var{c1}}\\ln\\left(\\frac{\\var{a1*r1} \\times 10^{\\var{b1}}}{\\var{a1} \\times 10^{\\var{b1}}}\\right)\\\\\n \n &=& \\frac{1}{\\var{c1}}\\ln(\\var{r1})\\\\\n \n &=&\\var{kvalue1}\n \n \\end{eqnarray*}\n \n \\] to 3 decimal places.

\n \n \n ", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\n \n \n

$\\rho_0=\\var{a} \\times 10^{\\var{b}},\\;\\;\\;\\;\\;\\rho(\\var{c})=\\var{a*r} \\times 10^{\\var{b}}$

\n \n \n \n

$k=\\;\\;$[[0]]

\n \n \n ", "gaps": [{"minvalue": "{kvalue}", "type": "numberentry", "maxvalue": "{kvalue}", "marks": 1.0, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}, {"prompt": "\n \n \n

$\\rho_0=\\var{a1} \\times 10^{\\var{b1}},\\;\\;\\;\\;\\;\\rho(\\var{c1})=\\var{a1*r1} \\times 10^{\\var{b1}}$

\n \n \n \n

$k=\\;\\;$[[0]]

\n \n \n ", "gaps": [{"minvalue": "{kvalue1}", "type": "numberentry", "maxvalue": "{kvalue1}", "marks": 1.0, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}], "statement": "\n

The size $\\rho(t)$ of a population satisfies:
\\[\\rho(t)=\\rho_0e^{kt}\\]
where $\\rho$ and $k$ are constants.

Given the following data, find $k$ in each case. Input your answers to 3 decimal places.

\n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(1..6)", "name": "a"}, "c": {"definition": "random(3..15)", "name": "c"}, "b": {"definition": "random(5..7)", "name": "b"}, "r1": {"definition": "if(r2=r,r+1,r)", "name": "r1"}, "r2": {"definition": "max(round(e^(k1*c1)),2)", "name": "r2"}, "k": {"definition": "random(0.005..0.15#0.005)", "name": "k"}, "a1": {"definition": "a+random(1..3)", "name": "a1"}, "k1": {"definition": "random(0.005..0.15#0.005)", "name": "k1"}, "r": {"definition": "max(round(e^(k*c)),2)", "name": "r"}, "b1": {"definition": "b+random(1,2)", "name": "b1"}, "kvalue1": {"definition": "precround(ln(r1)/c1,3)", "name": "kvalue1"}, "c1": {"definition": "c+random(1..3)", "name": "c1"}, "kvalue": {"definition": "precround(ln(r)/c,3)", "name": "kvalue"}}, "metadata": {"notes": "\n \t\t

2/07/2012:

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Added tags

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Forced answers to both parts to be exact to 3 decimal places.

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19/07/2012:

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Added description.

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Checked calculation.

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25/07/2012:

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Question appears to be working correctly.

\n \t\t", "description": "

Given $\\rho(t)=\\rho_0e^{kt}$, and values for $\\rho(t)$ for $t=t_1$ and a value for $\\rho_0$, find $k$. (Two examples).

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Logarithms: Solving equations 6", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["logarithmic equations", "logarithms", "logs", "quadratic equation", "solving a quadratic equation", "solving equations involving powers", "solving logarithmic equations"], "advice": "\n

In the equation
\\[\\var{c}\\left(\\var{a^2}^x \\right)+ \\var{d}\\left(\\var{a}^{x+1}\\right)=\\var{b}\\]

Let $y=\\var{a}^x$ and since $\\var{a^2}^x=(\\var{a}^2)^x=\\left(\\var{a}^x\\right)^2 $ on substitution this becomes:
\\[\\begin{eqnarray*} \\var{c}y^2 + \\var{d}\\left(\\var{a}y\\right)&=&\\var{b} \\\\ \\\\ \\Rightarrow \\var{c}y^2 + \\var{a*d}y&=&\\var{b}\\Rightarrow \\var{c}y^2 + \\var{a*d}y-\\var{b}=0\\\\ \\\\ \\Rightarrow (\\var{c}y+\\var{al})(y-\\var{abs(be)})&=&0 \\mbox{ on factorisation.} \\end{eqnarray*} \\]
This quadratic has solutions $\\displaystyle y=\\simplify[std]{{-al}/{c}},\\;\\;y=\\var{-be}$.

But $y=\\var{a}^x \\gt 0$ for all $x$ and so $\\displaystyle y=\\simplify[std]{{-al}/{c}}$ cannot be a solution for the original equation.

We are left with $y=\\var{-be}$ which gives:
\\[\\begin{eqnarray*} \\var{a}^x &=& \\var{-be}\\\\ \\Rightarrow x\\ln(\\var{a})&=&\\ln(\\var{-be})\\\\ \\Rightarrow x&=&\\frac{\\ln(\\var{-be})}{\\ln(\\var{a})} = \\var{ans1} \\end{eqnarray*} \\]

to 3 decimal places.

\n ", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"prompt": "\n

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

Input your answer to 3 decimal places.

\n ", "gaps": [{"minvalue": "ans1-tol", "type": "numberentry", "maxvalue": "ans1+tol", "marks": 2.0, "showPrecisionHint": false}], "type": "gapfill", "marks": 0.0}], "statement": "\n

Solve the following equation for $x$. Note that there is only one solution.

\\[\\var{c}\\left(\\var{a^2}^x \\right)+ \\var{d}\\left(\\var{a}^{x+1}\\right)=\\var{b}\\]

Hint: remember that $\\left(A^2\\right)^x=\\left(A^x\\right)^2$ for any number $A$.

\n ", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"a": {"definition": "random(4..9)", "name": "a"}, "be": {"definition": "random(-9..-2)", "name": "be"}, "c": {"definition": "random(2..6)", "name": "c"}, "b": {"definition": "-al*be", "name": "b"}, "d": {"definition": "if(c=2 or c=4,random(3,5,7,9),if(c=3,random(2,4,5,7,8),if(c=5,random(2,3,4,6,7),random(5,7,11))))", "name": "d"}, "ans1": {"definition": "precround(tans1,3)", "name": "ans1"}, "al": {"definition": "d*a-be*c", "name": "al"}, "tol": {"definition": 0.0, "name": "tol"}, "tans1": {"definition": "ln(-be)/ln(a)", "name": "tans1"}}, "metadata": {"notes": "

2/07/2012:

Added tags.

Added that the solution is to 3 decimal places.

Forced exact solution to 3 decimal places - no tolerance.

Improved display.

19/07/2012:

Added description.

Checked calculation.

New tolerance variable tol=0 for the answer.

25/07/2012:

Added tags.

Is it necessary to include the hint? It is a rather basic mathematical identity.

In the Advice section moved \\Rightarrow so that it is at the beginning of the line instead of the end of the previous line.

Question appears to be working correctly.

", "description": "

Solve for $x$: $c(a^2)^x + d(a)^{x+1}=b$ (there is only one solution for this example).

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}, {"name": "Simplify logarithms 1: (Video)", "extensions": [], "custom_part_types": [], "resources": [], "navigation": {"allowregen": true, "showfrontpage": false, "preventleave": false, "typeendtoleave": false}, "contributors": [{"name": "Bill Foster", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/6/"}], "functions": {}, "tags": ["log laws", "logarithm laws", "logarithmic expressions", "logarithms", "logs", "rules for logarithms", "simplifying logarithms", "video"], "advice": "\n \n \n

The rules for combining logs are

\n \n \n \n

\\[\\begin{eqnarray*}\n \n \\log_a(bc)&=&\\log_a(b)+\\log_a(c)\\\\\n \n \\\\\n \n \\log_a\\left(\\frac{b}{c}\\right)&=&\\log_a(b)-\\log_a(c)\\\\\n \n \\\\\n \n \\log_a(b^r)&=&r\\log_a(b)\n \n \\end{eqnarray*}\n \n \\]

\n \n \n \n

a)
Using these rules gives:
\\[ \\begin{eqnarray*}\n \n \\log_a(x^{\\var{a1}}y^{\\var{b1}})&=&\\log_a(x^{\\var{a1}})+\\log_a(y^{\\var{b1}})\\\\\n \n &=&\\var{a1}\\log_a(x)+\\var{b1}\\log_a(y)\n \n \\end{eqnarray*}\n \n \\]

\n \n \n \n

b)
\\[\\begin{eqnarray*}\n \n \\simplify[std]{{a2}/{b2}}\\log_a(x)+\\log_a(\\simplify{{c}*x+{d}})-\\log_a(\\simplify{x^(1/{b2})})&=&\\log_a(x^\\frac{\\var{a2}}{\\var{b2}})+\\log_a(\\simplify{{c}*x+{d}})-\\log_a(\\simplify{x^(1/{b2})})\\\\\n \n \\\\\n \n &=&\\log_a\\left(\\simplify[std]{(x^({a2}/{b2})*({c}x+{d}))/(x^(1/{b2}))}\\right)\\\\\n \n &=&\\log_a\\left(\\simplify{x^{f}*({c}x+{d})}\\right)\n \n \\end{eqnarray*}\n \n \\]

\n \n \n \n ", "rulesets": {"std": ["all", "fractionNumbers", "!collectNumbers", "!noLeadingMinus"]}, "parts": [{"stepspenalty": 0.0, "prompt": "

Express the following in terms of $\\log_a(x)$ and $\\log_a(y)$

\\[\\log_a(x^{\\var{a1}}y^{\\var{b1}})=\\alpha\\log_a(x)+\\beta\\log_a(y)\\]

$\\alpha=\\;\\;$[[0]], $\\beta=\\;\\;$[[1]]

In Show steps you will find a video explaining the rules of logarithms by going through simplification of logs of numbers rather than algebraic expressions.

", "gaps": [{"minvalue": "{a1}", "type": "numberentry", "maxvalue": "{a1}", "marks": 0.5, "showPrecisionHint": false}, {"minvalue": "{b1}", "type": "numberentry", "maxvalue": "{b1}", "marks": 0.5, "showPrecisionHint": false}], "steps": [{"prompt": "

This video covers the standard rules for logarithms.

These are, for any base $b \\gt 0$:

\\[\\begin{eqnarray*} &1.& \\log_b(p)+\\log_b(q)&=&\\log_b(p\\times q)\\\\&2.& \\log_b(p)-\\log_b(q)&=&\\log_b\\left(\\frac{p}{ q}\\right)\\\\ &3.& \\log_b(p^r)=r\\log_b(p)\\\\&4.& \\log_b(1)=0\\end{eqnarray*} \\]

", "type": "information", "marks": 0.0}], "marks": 0.0, "type": "gapfill"}, {"prompt": "\n

\\[\\simplify[std]{{a2}/{b2}}\\log_a(x)+\\log_a(\\simplify{{c}*x+{d}})-\\log_a(\\simplify{x^(1/{b2})})=\\log_a(q(x))\\]

$q(x)=\\;\\;$[[0]]

\n \n ", "gaps": [{"checkingaccuracy": 0.001, "vsetrange": [0.0, 1.0], "vsetrangepoints": 5.0, "checkingtype": "absdiff", "answersimplification": "std", "marks": 1.0, "answer": "((x ^ {f}) * (({c} * x) + {d}))", "type": "jme"}], "type": "gapfill", "marks": 0.0}], "statement": "

Answer the following questions on logarithms.

", "variable_groups": [], "progress": "ready", "type": "question", "variables": {"c": {"definition": "random(1..9)", "name": "c"}, "d": {"definition": "s4*random(2..9)", "name": "d"}, "f": {"definition": "precround((a2-1)/b2,0)", "name": "f"}, "s1": {"definition": "random(1,-1)", "name": "s1"}, "s4": {"definition": "random(1,-1)", "name": "s4"}, "a1": {"definition": "s1*random(2..9)", "name": "a1"}, "a2": {"definition": "1+b2*random(2..5)", "name": "a2"}, "b1": {"definition": "random(2..15)", "name": "b1"}, "b2": {"definition": "random(2..9)", "name": "b2"}}, "metadata": {"notes": "\n \t\t

2/06/2012:

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Changed statement to make question clearer.

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19/07/2012:

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25/07/2012:

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Question appears to be working correctly.

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17/08/2012:

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Made copy to include in Simplify Algebraic Expressions exam.

\n \t\t", "description": "

Express $\\log_a(x^{c}y^{d})$ in terms of $\\log_a(x)$ and $\\log_a(y)$. Find $q(x)$ such that $\\frac{f}{g}\\log_a(x)+\\log_a(rx+s)-\\log_a(x^{1/t})=\\log_a(q(x))$.

There is a video included explaining the rules of logarithms by going through simplification of logs of numbers rather than algebraic expressions.

", "licence": "Creative Commons Attribution 4.0 International"}, "showQuestionGroupNames": false, "question_groups": [{"name": "", "pickingStrategy": "all-ordered", "pickQuestions": 0, "questions": []}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}], "extensions": [], "custom_part_types": [], "resources": []}