Question

Am nevoie de ajutor ex 3 dau coronița ​

Answer 1

Răspuns:

Explicație pas cu pas:

$a) ~a=log_25.~~~log_{100}20=\frac{log_220}{log_2100}=\frac{log_2(2^2*5)}{log_2(2^2*5^2)}=\frac{log_22^2+log_25}{log_22^2+log_25^2}=\frac{2*log_22+a}{2*log_22+2*log_25}=\frac{2*1+a}{2*1+2*a}=\frac{2+a}{2+2a}\\ b)~a=log_345=log_3(3^2*5)=log_33^2+log_35=2*log_33+log_35=2*1+log_35=2+log_35.\\Deci,~log_35=a-2.\\log_{15}75=\frac{log_375}{log_315}=\frac{log_3(3*5^2)}{log_3(3*5)}=\frac{log_33+log_35^2}{log_33+log_35}=\frac{1+2*log_35}{1+a-2}=\frac{1+2(a-2)}{a-1}=\frac{2a-3}{a-1} .$