Question

sabendo que log2=0,30 log3= 0,47 e log 5 =0,69 e log7 = 0,85 calcule a) log15 b) log9 c) log1,25 d) log40 na base5

Answer 1

a)

$log\,15\\\\\\log\,(3~.~5)\\\\\\Utilizando~a~Propriedade~do~Logaritmo~de~um~Produto:\\\\\\log\,3~+~log\,5\\\\\\0,47~+~0,69\\\\\\\boxed{1,16}$

b)

$log\,9\\\\\\log\,(3~.~3)\\\\\\Utilizando~a~Propriedade~do~Logaritmo~de~um~Produto:\\\\\\log\,3~+~log\,3\\\\\\0,47~+~0,47\\\\\\\boxed{0,94}$

c)

$log\,1,25\\\\\\log\,\frac{125}{100}\\\\\\log\,\frac{5~.~5~.~5}{10~.~10}\\\\\\log\,\frac{5^3}{10^2}\\\\\\Utilizando~a~Propriedade~do~Logaritmo~de~um~Quociente:\\\\\\log\,5^3~-~log\,10^2\\\\\\Utilizando~a~Propriedade~do~Logaritmo~de~uma~Potencia:\\\\\\3.log\,5~-~2.log\,10\\\\\\3~.~(0,69)~-~2~.~(1)\\\\\\2,07~-~2\\\\\\\boxed{0,07}$

d)

$log_{_{5}}40\\\\\\Utilizando~a~Propriedade~da~Troca~de~Base:\\\\\\\frac{log_{_{10}}40}{log_{_{10}}5}\\\\\\\frac{log\,(2~.~2~.~2~.~5)}{log\,5}\\\\\\\frac{log\,(2^3~.~5)}{log\,5}\\\\\\Utilizando~a~Propriedade~do~Logaritmo~de~um~Produto:\\\\\\\frac{log\,2^3~+~log\,5}{log\,5}\\\\\\Utilizando~a~Propriedade~do~Logaritmo~de~um~Produto\\\\\\\frac{3.log\,2~+~log\,5}{log\,5}\\\\\\\frac{3~.~(0,30)~+~0,69}{0,69}\\\\\\\frac{0,90~+~0,69}{0,69}\\\\\\\frac{1,59}{0,69}~\approx~\boxed{2,30}$