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Solve equations involving logs and exponential functions, by using inverse operations.
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\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}\\]
\nTake logarithm base $\\var{A}$:
\n\\[\\simplify{x+{c}}=\\log_\\var{A}(\\var{B*D})=\\var{log(B*D)/log(A)}\\]
\nAdd/subtract $\\var{c}$:
\n\\[x=\\var{x}\\]
\nRounding to 2 decimal places gives $x=\\var{precround(x,2)}$
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", "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}\\]
\nSquare:
\n\\[\\simplify{exp({A}+x)}=\\var{D}^2=\\var{D^2}\\]
\nTake natural log:
\n\\[\\simplify{{A}+x}=\\ln(\\var{D^2})=\\var{ln(D^2)}\\]
\nSubtract $\\var{A}$:
\n\\[x=\\var{ln(D^2)}-\\var{A}=\\var{x}\\]
\nRounding to 2 decimal places gives $x=\\var{precround(x,2)}$
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", "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}\\]
\nDivide into $\\var{A}$:
\n\\[\\simplify{10^({B}*x-{C})}=\\frac{\\var{A}}{\\var{D}}=\\var{A/D}\\]
\nTake log base 10:
\n\\[\\simplify{{B}*x-{C}}=\\log(\\var{A/D})=\\var{log(A/D)}\\]
\nAdd $\\var{C}$:
\n\\[\\simplify{{B}*x}=\\var{log(A/D)}+\\var{C}=\\var{log(A/D)+C}\\]
\nDivide by $\\var{B}$:
\n\\[x=\\frac{\\var{log(A/D)+C}}{\\var{B}}=\\var{x}\\]
\nRounding to 2 decimal places gives $x=\\var{precround(x,2)}$
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", "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}\\]
\nSubtract from $\\var{A}$:
\n\\[\\simplify{ln((x+{B})/{C})}=\\var{A}-\\var{D}=\\var{A-D}\\]
\nTake $e$ to the power:
\n\\[\\simplify{(x+{B})/{C}}=e^\\var{A-D}=\\var{exp(A-D)}\\]
\nMultiply by $\\var{C}$:
\n\\[\\simplify{x+{B}}=\\var{C}\\times\\var{exp(A-D)}=\\var{C*exp(A-D)}\\]
\nAdd/subtract $\\var{abs(B)}$:
\n\\[x=\\var{x}\\]
\nRounding to 2 decimal places gives $x=\\var{precround(x,2)}$.
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", "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}\\]
\nDivide by $\\var{B}$:
\n\\[\\simplify{log(({A}-x)^2)}=\\frac{\\var{D}}{\\var{B}}=\\var{D/B}\\]
\nTake $10$ to the power:
\n\\[\\simplify{({A}-x)^2}=10^\\var{D/B}=\\var{10^(D/B)}\\]
\nSquare root:
\n\\[\\simplify{{A}-x}=\\pm\\sqrt{\\var{10^(D/B)}}=\\pm\\var{sqrt(10^(D/B))}\\]
\nSubtract from $\\var{A}$:
\n\\[x=\\var{A}\\mp\\var{sqrt(10^(D/B))}=\\var{x}\\text{ or }\\var{x2}\\]
\nRounding to 2 decimal places gives $x=\\var{precround(x,2)}$ or $x=\\var{precround(x2,2)}$
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