That equation is equivalent to \(\displaystyle \left( {\frac{5}{7}} \right)^x = \frac{7}{2}\) which has solution \(\displaystyle x = \frac{{\log (7) - \log (2)}}{{\log (5) - \log (7)}}\)

That equation is equivalent to \(\displaystyle \left( {\frac{5}{7}} \right)^x = \frac{7}{2}\) which has solution \(\displaystyle x = \frac{{\log (7) - \log (2)}}{{\log (5) - \log (7)}}\)

\(\displaystyle 2 \cdot 5^x = 3^{x + 1} \) is equivalent to \(\displaystyle \left( {\frac{5}{3}} \right)^x = \frac{3}{2}\) which has solution \(\displaystyle x = \frac{{\log (3) - \log (2)}}{{\log (5) - \log (3)}}\)
To see how that works divide both sides by \(\displaystyle 3^x\) and also by \(\displaystyle 2\).
Note that \(\displaystyle \frac{{5^x }}{{3^x }} = \left( {\frac{5}{3}} \right)^x \).

\(\displaystyle 2 \cdot 5^x = 3^{x + 1} \) is equivalent to \(\displaystyle \left( {\frac{5}{3}} \right)^x = \frac{3}{2}\) which has solution \(\displaystyle x = \frac{{\log (3) - \log (2)}}{{\log (5) - \log (3)}}\)
To see how that works divide both sides by \(\displaystyle 3^x\) and also by \(\displaystyle 2\).
Note that \(\displaystyle \frac{{5^x }}{{3^x }} = \left( {\frac{5}{3}} \right)^x \).