Two minus log 3X equals log (X over 12)

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Answer to a math question Two minus log 3X equals log (X over 12)

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Hermann

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$\log_{ 10 }({ {10}^{2} })-\log_{ 10 }({ 3x })=\log_{ 10 }({ \frac{ x }{ 12 } })$

$\log_{ 10 }({ \frac{ {10}^{2} }{ 3x } })=\log_{ 10 }({ \frac{ x }{ 12 } })$

$\log_{ 10 }({ \frac{ 100 }{ 3x } })=\log_{ 10 }({ \frac{ x }{ 12 } })$

${10}^{\log_{ 10 }({ \frac{ 100 }{ 3x } })}={10}^{\log_{ 10 }({ \frac{ x }{ 12 } })}$

$\frac{ 100 }{ 3x }={10}^{\log_{ 10 }({ \frac{ x }{ 12 } })}$

$\frac{ 100 }{ 3x }=\frac{ x }{ 12 }$

$\begin{array} { l }x=-20,\\x=20\end{array}$

$\begin{array} { l }2-\log_{ 10 }({ 3 \times \left( -20 \right) })=\log_{ 10 }({ \frac{ -20 }{ 12 } }),\\x=20\end{array}$

$\begin{array} { l }2-\log_{ 10 }({ 3 \times \left( -20 \right) })=\log_{ 10 }({ \frac{ -20 }{ 12 } }),\\2-\log_{ 10 }({ 3 \times 20 })=\log_{ 10 }({ \frac{ 20 }{ 12 } })\end{array}$

$\begin{array} { l }\textnormal{Undefined},\\2-\log_{ 10 }({ 3 \times 20 })=\log_{ 10 }({ \frac{ 20 }{ 12 } })\end{array}$

$\begin{array} { l }\textnormal{Undefined},\\0.221849=0.221849\end{array}$

$\begin{array} { l }x≠-20,\\0.221849=0.221849\end{array}$

$\begin{array} { l }x≠-20,\\x=20\end{array}$

$x=20$

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