// Numbas version: exam_results_page_options {"name": "Equations involving logs", "metadata": {"description": "

Solve equations involving logs and exponential functions, by using inverse operations.

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Solve equation involving exponential function.

", "licence": "Creative Commons Attribution-ShareAlike 4.0 International"}, "statement": "

Solve the equation

\n

\\[\\simplify{{A}^(x+{c})/{B}}=\\var{D}\\]

", "advice": "

Multiply by $\\var{B}$:

\n

\\[\\simplify{{A}^(x+{c})}=\\var{B}\\times\\var{D}=\\var{B*D}\\]

\n

Take logarithm base $\\var{A}$:

\n

\\[\\simplify{x+{c}}=\\log_\\var{A}(\\var{B*D})=\\var{log(B*D)/log(A)}\\]

\n

Add/subtract $\\var{c}$:

\n

\\[x=\\var{x}\\]

\n

Rounding to 2 decimal places gives $x=\\var{precround(x,2)}$

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

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Solve equation involving exponential function

", "licence": "Creative Commons Attribution-ShareAlike 4.0 International"}, "statement": "

Solve the equation

\n

\\[\\simplify{sqrt(exp({A}+x))}=\\var{D}\\]

", "advice": "

Starting from the equation:

\n

\\[\\simplify{sqrt(exp({A}+x))}=\\var{D}\\]

\n

Square:

\n

\\[\\simplify{exp({A}+x)}=\\var{D}^2=\\var{D^2}\\]

\n

Take natural log:

\n

\\[\\simplify{{A}+x}=\\ln(\\var{D^2})=\\var{ln(D^2)}\\]

\n

Subtract $\\var{A}$:

\n

\\[x=\\var{ln(D^2)}-\\var{A}=\\var{x}\\]

\n

Rounding to 2 decimal places gives $x=\\var{precround(x,2)}$

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

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Solve equation involving exponential function.

", "licence": "Creative Commons Attribution-ShareAlike 4.0 International"}, "statement": "

Solve the equation

\n

\\[\\simplify{{A}/10^({B}*x-{C})}=\\var{D}\\]

", "advice": "

Starting from the equation:

\n

\\[\\simplify{{A}/10^({B}*x-{C})}=\\var{D}\\]

\n

Divide into $\\var{A}$:

\n

\\[\\simplify{10^({B}*x-{C})}=\\frac{\\var{A}}{\\var{D}}=\\var{A/D}\\]

\n

Take log base 10:

\n

\\[\\simplify{{B}*x-{C}}=\\log(\\var{A/D})=\\var{log(A/D)}\\]

\n

Add $\\var{C}$:

\n

\\[\\simplify{{B}*x}=\\var{log(A/D)}+\\var{C}=\\var{log(A/D)+C}\\]

\n

Divide by $\\var{B}$:

\n

\\[x=\\frac{\\var{log(A/D)+C}}{\\var{B}}=\\var{x}\\]

\n

Rounding to 2 decimal places gives $x=\\var{precround(x,2)}$

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

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Solve equation involving logarithmic function.

", "licence": "Creative Commons Attribution-ShareAlike 4.0 International"}, "statement": "

Solve the equation

\n

\\[\\simplify{{A}-ln((x+{B})/{C})}=\\var{D}\\]

", "advice": "

Starting from the equation:

\n

\\[\\simplify{{A}-ln((x+{B})/{C})}=\\var{D}\\]

\n

Subtract from $\\var{A}$:

\n

\\[\\simplify{ln((x+{B})/{C})}=\\var{A}-\\var{D}=\\var{A-D}\\]

\n

Take $e$ to the power:

\n

\\[\\simplify{(x+{B})/{C}}=e^\\var{A-D}=\\var{exp(A-D)}\\]

\n

Multiply by $\\var{C}$:

\n

\\[\\simplify{x+{B}}=\\var{C}\\times\\var{exp(A-D)}=\\var{C*exp(A-D)}\\]

\n

Add/subtract $\\var{abs(B)}$:

\n

\\[x=\\var{x}\\]

\n

Rounding to 2 decimal places gives $x=\\var{precround(x,2)}$.

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

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Solve equation involving logarithmic function.

", "licence": "Creative Commons Attribution-ShareAlike 4.0 International"}, "statement": "

Solve the equation

\n

\\[\\simplify{{B}*log(({A}-x)^2)}=\\var{D}\\]

", "advice": "

Starting from the equation

\n

\\[\\simplify{{B}*log(({A}-x)^2)}=\\var{D}\\]

\n

Divide by $\\var{B}$:

\n

\\[\\simplify{log(({A}-x)^2)}=\\frac{\\var{D}}{\\var{B}}=\\var{D/B}\\]

\n

Take $10$ to the power:

\n

\\[\\simplify{({A}-x)^2}=10^\\var{D/B}=\\var{10^(D/B)}\\]

\n

Square root:

\n

\\[\\simplify{{A}-x}=\\pm\\sqrt{\\var{10^(D/B)}}=\\pm\\var{sqrt(10^(D/B))}\\]

\n

Subtract from $\\var{A}$:

\n

\\[x=\\var{A}\\mp\\var{sqrt(10^(D/B))}=\\var{x}\\text{ or }\\var{x2}\\]

\n

Rounding to 2 decimal places gives $x=\\var{precround(x,2)}$ or $x=\\var{precround(x2,2)}$

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$x=\\;$[[0]] or $x=\\;$[[1]]

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