Question

Log2 (3-x)=log (3-x)

Answer 1

Fazendo a mudança de base para a base e, ficamos com o seguinte:

$\log _{2} (3-x) = \frac{\ln _{} (3-x)}{\ln _{} (2)}$

$\log _{10} (3-x) = \frac{\ln _{} (3-x)}{\ln _{} (10)}$

Nisso, ficamos com:

$\frac{\ln _{} (3-x)}{\ln _{} (2)} = \frac{\ln _{} (3-x)}{\ln _{} (10)}$

Que isolando o x, chega a resposta:

$\frac{\ln _{} (3-x)}{\ln _{} (2)} = \frac{\ln _{} (3-x)}{\ln _{} (10)} \\ \\ \frac{\ln _{} (3-x)}{\ln _{} (2)} - \frac{\ln _{} (3-x)}{\ln _{} (10)} = 0 \\ \\ \ln _{} (3-x)( \frac{1}{\ln _{} (2)} -\frac{1}{\ln _{} (10)})=0 \\ \\ \ln _{} (3-x)=0 \ ou \ ( \frac{1}{\ln _{} (2)} -\frac{1}{\ln _{} (10)})=0 \\ \\ Mas, \ ( \frac{1}{\ln _{} (2)} -\frac{1}{\ln _{} (10)}) \neq 0 \\ \\ Logo, \ \ln _{} (3-x)=0 \\ \\ e^{0} =3-x \\ \\ 1 = 3-x \\ \\ x = 2$