Question

Encontrar um número x>0 tal que: Log5 x + Log5 2 = 2

Answer 1

Propriedade: logaritmo do produto

$\fbox{ \ell og_{\alpha}(\beta\cdot\gamma)=\ell og_{\alpha}(\beta)+\ell og_{\alpha}(\gamma) }~~~~(0\ \textless \ \alpha \neq 1)~e~(0<\beta)$

Portanto

$\ell og_5(x)+\ell og_5(2)=2\\\\\ell og_5(x\cdot2)=2$

Aplicando a definição de logaritmo: 

$\fbox{ \ell og_{\alpha}(\beta)=y~\Leftrightarrow~\alpha^{y}=\beta }~~~~(0\ \textless \ \alpha \neq 1)~e~(0<\beta)$

Assim

$\ell og_5(x\cdot2)=2~\Leftrightarrow~5^2=2x\\\\25=2x\\\\\fbox{ x=\dfrac{25}{2} }$