[tex]~~~~~~~x\log (7) = (x+4) \log (3)\\\\\\\implies x \log(7) = x\log (3)+4 \log(3)\\\\\\\implies x \log(7) - x \log(3) = 4 \log(3)\\\\\\\implies x ( \log 7 - \log 3) = 4\log 3\\\\\\\implies x = \dfrac{4\log 3}{\log 7 - \log 3} \\\\\\\implies x \approx5.19[/tex]