Answer:

[tex]\qquad\qquad\boxed{ \sf{ \: \bf \: x = \dfrac{2log5 - log7}{log5 + 3log7} \: }}\\ \\ [/tex]

Step-by-step explanation:

Given equation is

[tex]\sf \: {5}^{2 - x} = {7}^{3x + 1} \\ \\ [/tex]

On taking log on both sides, we get

[tex]\sf \: log\left({5}^{2 - x} \right)= log\left({7}^{3x + 1}\right) \\ \\ [/tex]

[tex]\sf \:(2 - x)log5 = (3x + 1)log7 \\ \\ [/tex]

[tex]\qquad\boxed{ \sf{ \: \because \: log( {x}^{y}) = ylogx \: }} \\ \\ [/tex]

[tex]\sf \:2log5 - xlog5 = 3xlog7 + log7 \\ \\ [/tex]

[tex]\sf \:2log5 - log7 = xlog5 + 3xlog7 \\ \\ [/tex]

[tex]\sf \:2log5 - log7 = x(log5 + 3log7) \\ \\ [/tex]

[tex]\sf\implies \bf \: x = \dfrac{2log5 - log7}{log5 + 3log7} \\ \\ [/tex]

[tex]\rule{190pt}{2pt}[/tex]

Additional Information

[tex]\begin{gathered}\: \: \: \: \: \: \begin{gathered}\begin{gathered} \footnotesize{\boxed{ \begin{array}{cc} \small\underline{\frak{\pmb{{More \: Formulae}}}} \\ \\ \bigstar \: \bf{ log_{x}(x) = 1}\\ \\ \bigstar \: \bf{ log_{x}( {x}^{y} ) = y}\\ \\ \bigstar \: \bf{ log_{ {x}^{z} }( {x}^{w} ) = \dfrac{w}{z} }\\ \\ \bigstar \: \bf{ log_{a}(b) = \dfrac{logb}{loga} }\\ \\ \bigstar \: \bf{ {e}^{logx} = x}\\ \\ \bigstar \: \bf{ {e}^{ylogx} = {x}^{y}}\\ \\ \bigstar \: \bf{log1 = 0}\\ \\ \end{array} }}\end{gathered}\end{gathered}\end{gathered}[/tex]