Question

Dados log 2=0,30 e log 3= 0,48, quanto vale: a) Log 500 b) Log 72 c) Log 7,5 d) Log 0,3

Answer 1

Nesta questão, vamos utilizar as propriedades de logaritmos e, também, a fatoração.

a)

$\underline{Fatorando}~o~logaritmando~500\\\\\\\log500~=~\log\left(2\cdot2\cdot5\cdot5\cdot5\right)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}\\\\\\\log500~=~\log2~+~\log2~+~\log5~+~\log5~+~\log5\\\\\\\log500~=~2\cdot\log2~+~3\cdot\log5\\\\\\Note~que~nao~temos~o~valor~para~\log5.\\No~entanto,~sabemos~o~valor~de~\log10~e~de~\log2\\Vamos~entao~reescrever~5~como~\frac{10}{2}$

$\log500~=~2\cdot\log2~+~3\cdot\log\left(\dfrac{10}{2}\right)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~quociente}\\\\\\\log500~=~2\cdot\log2~+~3\cdot\left(\log10~-~\log2\right)\\\\\\Substituindo~os~valores\\\\\\\log500~=~2\cdot0,30~+~3\cdot(1~-~0,30)\\\\\\\log500~=~0,60~+~3\cdot0,70\\\\\\\log500~=~0,60~+~2,10\\\\\\\boxed{\log500~=~2,70}$

b)

Da mesma forma que fizemos na (a), vamos fatorar o logaritmando 72 e aplicar as propriedades.

$\log72~=~\log\,(2\cdot2\cdot2\cdot3\cdot3)\\\\\\\log72~=~\log2~+~\log2~+~\log2~+~\log3~+~\log3\\\\\\\log72~=~3\cdot\log2~+~2\cdot\log3\\\\\\\log72~=~3\cdot0,30~+~2\cdot0,48\\\\\\\log72~=~0,90~+~0,96\\\\\\\boxed{\log72~=~1,86}$

c)

Note que 7,5 não é um numero inteiro, logo vamos reescreve-lo para podermos, posteriormente, fatora-lo.

$\log7,5~=~\log\left(\dfrac{75}{10}\right)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~quociente}\\\\\\\log7,5~=~\log75~-~\log10\\\\\\\underline{Fatorando}~o~logaritmando~75\\\\\\\log7,5~=~\log(3\cdot5\cdot5)~-~\log10\\\\\\Novamente,~nao~temos~\log5,~mas~temos~\log10~e~\log2\\\\\\log7,5~=~\log\left(3\cdot\dfrac{10}{2}\cdot\dfrac{10}{2}\right)~-~\log10$

$Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}\\\\\\log7,5~=~\log3~+~\log\dfrac{10}{2}~+~\log\dfrac{10}{2}\right)~-~\log10\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~quociente}\\\\\\log7,5~=~\log3~+~\left(\log10-\log2\right)+\left(\log10-\log2\right)~-~\log10\\\\\\Substituindo~os~valores\\\\\\\log7,5~=~0,48~+~(1-0,30)~+~(1-0,30)~-~1\\\\\\\log7,5~=~0,48~+~0,70~+~0,70~-~1\\\\\\\boxed{\log7,5~=~0,88}$

d)

Semelhante ao que foi feito em (c), vamos reescrever o logaritmando e aplicar as propriedades.

$\log0,3~=~\log\left(\dfrac{3}{10}\right)\\\\\\\log0,3~=~\log3~-~\log10\\\\\\\log0,3~=~0,48~-~1\\\\\\\boxed{\log0,3~=\,-0,52}$