Question

Não estou conseguindo resolver o e-) e o f-) (logaritmo)

Answer 1

Sabendo que log 2 = 0,3 e log 3 = 0,47, temos:

$Resolu\c{c}\~ao \to \left\{\begin{array}{ccc}log\ 72 = x \\\\log\ 6^2*2 = x \\\\log\ 2^2 * 3^2 * 2 = x \\\\log\ 2^2 + log\ 3^2 + log\ 2 = x \\\\2*log\ 2 + 2*log\ 3 + log\ 2 = x \\\\2*0,3 + 2*0,47 + 0,3 = x \\\\0,6 +0,94 + 0,3 = x \\\\ \boxed{\boxed{1,84 = x }}\end{array}\right$

$Resolu\c{c}\~ao \to \left\{\begin{array}{ccc}log\ 3600 = x \\\\log\ 36*100 = x \\\\log\ 6^2*10^2 = x \\\\log\ 2^2*3^2*2^2*5^2 = x \\\\log\ 2^2 + log\ 3^2 + log\ 2^2+ log\ 5^2 = x \\\\2*log\ 2+2*log\ 3+2*log\ 2+2*log\ 5 = x \\\\2*0,3+2*0,47+2*0,3+2*(1-0,3) = x \\\\0,6+ 0,94+0,6+1,4 = x\\\\\boxed{\boxed{3,54 = x }}\end{array}\right$

Espero ter ajudado. =^.^=