Question

(ESPM-SP) Sendo log 2 = a e log 3 = b, o valor do
$log_{9}(160)$ é igual a:


a) 4a + b / 2
b) 4a + 1 / 2b
c) 2a + 3b / 2
d) 4b + 2 / a
e) a + 1 / 3b​

Answer 1

Resposta:

log₉ (160) = log 160/log 9

=log (2⁵ *5) /log 9

=(log 2⁵+log 5)/ log 3²

=[5log 2 +(log 10/2)]/(2 log 3)

=[5 log 2 +log 10- log 2] /(2 log 3)

=[5a+1-a]/(2b)

=[4a+1]/2b

Letra B