Solve for $x$.
\n$\\log_\\var{b1}(x-\\var{b2})-\\log_\\var{b1}\\left(\\displaystyle\\frac{1}{x}\\right)=\\var{b4}$
\n$x=$ [[0]]
", "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "showCorrectAnswer": true, "marks": 0, "variableReplacementStrategy": "originalfirst"}, {"gaps": [{"checkingaccuracy": 0.001, "checkvariablenames": false, "variableReplacements": [], "marks": "2", "vsetrangepoints": 5, "showpreview": true, "expectedvariablenames": [], "answer": "e^({m}/{p})/{v}", "checkingtype": "absdiff", "scripts": {}, "showFeedbackIcon": true, "vsetrange": [0, 1], "notallowed": {"message": "", "partialCredit": 0, "showStrings": false, "strings": ["*", ")^"]}, "type": "jme", "showCorrectAnswer": true, "variableReplacementStrategy": "originalfirst"}], "type": "gapfill", "prompt": "Solve for $x$ and leave your answer in the form $x=\\displaystyle\\frac{e^{a}}{b}$.
\n$\\var{p}\\ln(x)+\\ln(\\var{q})=\\var{m}$
\n$x=$ [[0]]
", "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "showCorrectAnswer": true, "marks": 0, "steps": [{"type": "information", "prompt": "You may find the following conversion useful
\n\\[\\ln(x)=\\log_e(x)\\]
", "showFeedbackIcon": true, "scripts": {}, "variableReplacements": [], "showCorrectAnswer": true, "marks": 0, "variableReplacementStrategy": "originalfirst"}], "stepsPenalty": 0, "variableReplacementStrategy": "originalfirst"}], "advice": "We can use the logarithm law
\n\\[k\\log_a(x)=\\log_a(x^k)\\text{,}\\]
\nto also give a more specific rule
\n\\[\\begin{align}
\\log_a\\left(\\frac{1}{x}\\right)&=\\log_a(x^{-1})\\\\
&=-\\log_a(x)\\text{.}
\\end{align}\\]
This means we can write our expression as
\n\\[\\log_\\var{b1}(x-\\var{b2})+\\log_\\var{b1}({x})=\\var{b4}\\text{.}\\]
\nThen using the rule
\n\\[\\log_a(x)+\\log_a(y)=\\log_a(x\\times y)\\text{,}\\]
\nwe can write our equation as
\n\\[\\begin{align}
\\log_\\var{b1}(x(x-\\var{b2}))&=\\var{b4}\\\\
\\log_\\var{b1}(x^2-\\var{b2}x)&=\\var{b4}\\text{.}\\\\
\\end{align}\\]
We then rely on the definition of $\\log_a$
\n\\[b=a^c \\Longleftrightarrow \\log_{a}b=c\\]
\nto write our equation as
\n\\[\\begin{align}
x^2-\\var{b2}x&=\\var{b1}^\\var{b4}\\\\
&=\\var{b1^b4}\\text{.}
\\end{align}\\]
We can then write out our equation and solve either by factorising or using the quadratic formula;
\n\\[\\begin{align}
x^2-\\var{b2}x-\\var{b1^{b4}}&=0\\\\
(x+2)(x-\\var{b})&=0\\text{.}
\\end{align}\\]
As logarithms can only be applied to positive numbers, the only possible value for $x$ is $\\var{b}$.
\n$\\ln(x)$ is a shorthand for $\\log_e(x)$, so we can apply the same laws of logarithms here.
\nTherefore applying the rule
\n\\[k\\log_a(x)=\\log_a(x^k)\\]
\nwe can write our equation as
\n\\[\\ln(x^\\var{p})+\\ln(\\var{q})=\\var{m}\\text{.}\\]
\nThen using the rule
\n\\[\\log_a(x)+\\log_a(y)=\\log_a(x\\times y)\\]
\nwe can write our equation as
\n\\[\\ln(\\var{q}x^\\var{p})=\\var{m}\\text{.}\\]
\nAs $\\ln=\\log_e$ we can use
\n\\[a=b^c \\Longleftrightarrow \\log_ba=c\\]
\nto write our equation as
\n\\[\\var{q}x^\\var{p}=e^\\var{m}\\text{.}\\]
\nWe then just need to rearrange our equation
\n\\[\\begin{align}
\\var{q}x^\\var{p}&=e^\\var{m}\\\\[0.5em]
x^\\var{p}&=\\frac{e^\\var{m}}{\\var{q}}\\\\[0.5em]
x&=\\frac{e^{\\var{m}/\\var{p}}}{\\var{q^(1/{p})}}
\\end{align}\\]
Apply and combine logarithm laws in a given equation to find the value of $x$.
"}, "ungrouped_variables": ["x1", "y1", "z1", "b1", "c", "b4", "b", "b2"], "preamble": {"css": "", "js": ""}, "extensions": [], "variables": {"m": {"name": "m", "description": "", "definition": "random(2..20)", "group": "part c", "templateType": "anything"}, "b1": {"name": "b1", "description": "", "definition": "random(2..8 #2)", "group": "Ungrouped variables", "templateType": "anything"}, "y1": {"name": "y1", "description": "", "definition": "random(2..6)", "group": "Ungrouped variables", "templateType": "anything"}, "z1": {"name": "z1", "description": "", "definition": "random(2..6 except y1)", "group": "Ungrouped variables", "templateType": "anything"}, "q": {"name": "q", "description": "", "definition": "v^p", "group": "part c", "templateType": "anything"}, "p": {"name": "p", "description": "", "definition": "random(3..6)", "group": "part c", "templateType": "anything"}, "b4": {"name": "b4", "description": "", "definition": "random(2..4 except b1)", "group": "Ungrouped variables", "templateType": "anything"}, "v": {"name": "v", "description": "", "definition": "random(2..10)", "group": "part c", "templateType": "anything"}, "b": {"name": "b", "description": "", "definition": "c/2", "group": "Ungrouped variables", "templateType": "anything"}, "x1": {"name": "x1", "description": "", "definition": "repeat(random(2..20),8)", "group": "Ungrouped variables", "templateType": "anything"}, "b2": {"name": "b2", "description": "", "definition": "b-2", "group": "Ungrouped variables", "templateType": "anything"}, "c": {"name": "c", "description": "", "definition": "b1^b4", "group": "Ungrouped variables", "templateType": "anything"}}, "rulesets": {}, "statement": "", "variable_groups": [{"variables": ["p", "v", "q", "m"], "name": "part c"}], "type": "question", "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "David Martin", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/130/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}]}]}], "contributors": [{"name": "Christian Lawson-Perfect", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/7/"}, {"name": "David Martin", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/130/"}, {"name": "Hannah Aldous", "profile_url": "https://numbas.mathcentre.ac.uk/accounts/profile/1594/"}]}