Question

Qual o valor de x na equação

$3 (x+1)= 3.log~4+ 2.log \sqrt{5}-log~32$

Answer 1

Olá Mazzaropii!!

$\\ \displaystyle \mathsf{3(x + 1) = 3 \cdot \log 4 + 2 \cdot \log \sqrt{5} - \log 32} \\\\ \mathsf{3(x + 1) = \log 4^3 + \log (\sqrt{5})^2 - \log 32} \\\\ \mathsf{3(x + 1) = (\log 64 - \log 32) + \log 5} \\\\ \mathsf{3(x + 1) = \log \left ( \frac{64}{32} \right ) + \log 5} \\\\ \mathsf{3(x + 1) = \log 2 + \log 5} \\\\ \mathsf{3(x + 1) = \log (2 \cdot 5)} \\\\ \mathsf{3(x + 1) = \log 10} \\\\ \mathsf{3x + 3 = \log 10^1} \\\\ \mathsf{3x = 1 - 3} \\\\ \boxed{\mathsf{x = - \frac{2}{3}}}$