Question

como se resolve: (log d)² + log d² + 1 = 0 ??

Answer 1

Pela propriedade logaritmo da potência:

$\log d^2=2\cdot\log d$

Ficando assim:

$(\log d)^2+2\cdot\log d+1=0$

Substituindo (log d) por x:

$x^2+2x+1=0$

Resolvendo por Bhaskara:

$\Delta=b^2-4ac\\ \\\Delta=2^2-4\cdot1\cdot1\\ \\\Delta=4-4\\ \\\Delta=0$

$x=\frac{-2\pm\sqrt{0}}{2\cdot1}\\ \\x=-\frac{2}{2}\\ \\x=-1$

Portanto:

$\log d=-1\\ \\d=10^{-1}\\ \\d=0,1$

Answer 2

Olá!

$\\ \displaystyle \mathsf{(\log d)^2 + \log d^2 + 1 = 0} \\\\ \mathsf{(\log d)^2 + 2 \cdot \log d + 1 = 0} \\\\ \mathsf{(\log d)^2 + 2 \cdot \log d \cdot 1 + 1^2 = 0} \\\\ \mathsf{(\log d + 1)^2 = 0} \\\\ \mathsf{\log d + 1 = \sqrt{0}} \\\\ \mathsf{\log d + 1 = 0} \\\\ \mathsf{\log d = - 1} \\\\ \mathsf{10^{- 1} = d} \\\\ \boxed{\mathsf{d = \frac{1}{10}}}$

Obs.:

$\mathbf{\log a^c = c \cdot \log a}$