f(x) = log de (x+1) na base 3,
a) f(2) = log(2+1) = log3 = 1
f) f(0) =log(0+1) = log 1 3^x = 1 ==. 3^x = 3^0 ==> x = 0
f(26) = log(26+1) = log 27 3^x = 27 ==. 3^x = 3^3 ==> x = 3
g(x) = 4+log de x na base 2
b) g(2) = 4 +log2 = 4+1 = 5
g(1)= 4+ log 1 = 4+ log 1 ==> 4 + 0 = 4 2^x = 1 ==. 2^x = 2^0 ==> x = 0
g(raiz de2)= 4+ logV2 = 4 + 1/2 = (8+1)/2 = 9/2 2^x =V2 ==>2^x = 2^1/2 ==>x = 1/2
h(x) = log de 2x,de
c) h(5) = log 2.5 = log10 = 1
