Resposta :
Hola.
Vou fazer o mais complicado:
b) [tex]log_(625) \sqrt{5} = log_5^{4} ~5^ \frac{1}{2} = \frac{1}{4}* \frac{1}{2} *log_55 = \frac{1}{4} * \frac{1}{2}*1 = \frac{1}{8} [/tex]
c) log5 (o,oooo64) = log_5 0,2^6 = 6*log_5 0,2 = 6*log_5 2/10 = 6*log_ 1/2 =
6*(log_5 1 - log_5 5) = 6*(0 - 1) = -6
Pois não:
d) Log49 3√7 = log_49 √(3²*7) = log_49 √(9*7) = log_7² √(9*7) = 1/2 * log_7√(9*7) = 1/2 * (log_7 √9 + log_7√7 = 1/2 *(log_7 3 + log_7 7^1/2 = 1/2*(log_7 3 + 1/2*log_7 7) = 1/2*(log_7 3 + 1/2 * 1) = 1/2*(log_7 3 + 1/2) = 1/2*log_7 2 + 1/2 * 1/2 = 1/2*log_7 3 + 1/4 Ou
log_7√3 + 1/4
Vou fazer o mais complicado:
b) [tex]log_(625) \sqrt{5} = log_5^{4} ~5^ \frac{1}{2} = \frac{1}{4}* \frac{1}{2} *log_55 = \frac{1}{4} * \frac{1}{2}*1 = \frac{1}{8} [/tex]
c) log5 (o,oooo64) = log_5 0,2^6 = 6*log_5 0,2 = 6*log_5 2/10 = 6*log_ 1/2 =
6*(log_5 1 - log_5 5) = 6*(0 - 1) = -6
Pois não:
d) Log49 3√7 = log_49 √(3²*7) = log_49 √(9*7) = log_7² √(9*7) = 1/2 * log_7√(9*7) = 1/2 * (log_7 √9 + log_7√7 = 1/2 *(log_7 3 + log_7 7^1/2 = 1/2*(log_7 3 + 1/2*log_7 7) = 1/2*(log_7 3 + 1/2 * 1) = 1/2*(log_7 3 + 1/2) = 1/2*log_7 2 + 1/2 * 1/2 = 1/2*log_7 3 + 1/4 Ou
log_7√3 + 1/4