Verified answer

[tex]\Large\boxed{\begin{array}{l}\underline{\rm De\!\!~finic_{\!\!,}\tilde ao\,de\,logaritmo}\\\sf \log_ba=x\iff b^x=a\begin{cases}\sf a > 0\\\sf b > 0\\\sf b\ne1\end{cases}\\\underline{\rm Logaritmo\,do\,produto}\\\sf \log_c(a\cdot b)=\log_ca+\log_cb\\\underline{\rm Logaritmo\,do\,quociente}\\\sf \log_c\bigg(\dfrac{a}{b}\bigg)=\log_c\!a-\log_c\!b\end{array}}[/tex]

[tex]\Large\boxed{\begin{array}{l}\sf \log x+\log 2+\log3-\log 5=2\\\sf \log\bigg(\dfrac{x\cdot2\cdot3}{5}\bigg)=2\\\\\sf\log\bigg(\dfrac{6x}{5}\bigg)=2\\\\\sf \dfrac{6x}{5}=10^2\\\\\sf 6x=5\cdot100\\\sf 6x=500\\\sf x=\dfrac{500\div2}{6\div2}\\\\\sf x=\dfrac{250}{3}\end{array}}[/tex]