Question

Se 3^m=a e 3^n=b, a >0 e b>0, então qual o valor de:
${3}^{\frac{m - 2n}{2} }$
​

Answer 1

Resposta:

Letra D

Explicação passo a passo:

Algumas propriedades dos logaritmos

$1)a^{log_ab=b$

$2)log_ab^x=x.log_ab\\\\3)log_ab=\frac{log_cb}{log_ca}$

$3^m=a~~e~~3^n=b\\\\\ log3^m=loga\implies~~m.log3=loga\impliesm\implies m=\frac{loga}{log3}\implies m=log_3a \\\\log3^n=logb\implies~~n.log3=logb\impliesm\implies n=\frac{logb}{log3}\implies n=log_3b$

$3^{\frac{m-2n}{2} } =3^{\frac{log_3a-2.log_3b}{2} } =3^{\frac{log_3a-log_3b^2}{2} } =3^{\frac{1}{2}(log_3\frac{^a}{b^2}) } =3^{log_3}^{(\frac{a}{b^2})^{\frac{1}{2} } } =(\frac{a}{b^2})^{\frac{1}{2} } =\sqrt{\frac{a}{b^2}} =\frac{\sqrt{a} }{b}$