Question

Sendo logx a=6 logx b=4 e logx c=2, calcule log c raiz abc

Answer 1

Se a questão for calcular $log_c{ \sqrt{abc} }$
Então temos:

$log_c{ \sqrt{abc} }=log_c{(abc)^{\frac{1}{2}} }= \frac{1}{2}.log_cabc=\frac{1}{2}.(log_cabc)=\\\\ \frac{1}{2}.(log_ca +\log_cb+log_cc)=\\\\ =\frac{1}{2}.(\frac{log_xa}{log_xc}+\frac{log_xb}{log_xc}+1})\\\\ =\frac{1}{2}.(\frac{6}{2}+\frac{4}{2}+1})\\\\ =\frac{1}{2}.(3+2+1)=\frac{1}{2}.(6)=3$