Question

Dados log 2 = 0,301 e log 3 = 0,477, calcule :
a) log $\sqrt[5]{7,2}$=
b) log 14,4=
c) log 15 =

Answer 1

Neste exercício, utilizaremos, basicamente, fatoração e propriedades logarítmicas para reescrever os logaritmos dados em função de termos conhecidos.

Sendo assim, vamos começar relembrando algumas dessas propriedades.

$\sf Propriedade~do~Logaritmo~do~Produto:~~\boxed{\sf \log_b(a\cdot c)=\log_ba+\log_bc}\\\\Propriedade~do~Logaritmo~do~Quociente:~~\boxed{\sf \log_b\left(\dfrac{a}{c}\right)=\log_ba-\log_bc}\\\\Propriedade~do~Logaritmo~da~Potencia:~~\boxed{\sf \log_ba^c=c\cdot \log_ba}$

a)

$\sf Reescrevendo~o~radical~do~logaritmando~como~uma~\underline{potencia}\\\underline{de~expoente~fracionario}:\\\\\\\log\sqrt[5]{7,2}~=~\log7,2^{\frac{1}{5}}\\\\\\Agora,~reescrevendo~7,2~na~sua~\underline{forma~fracionaria}:\\\\\log\sqrt[5]{7,2}~=~\log\left(\dfrac{72}{10}\right)^{\frac{1}{5}}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~da~potencia}:\\\\\\\log\sqrt[5]{7,2}~=~\dfrac{1}{5}\cdot \log\left(\dfrac{72}{10}\right)$

$\sf Aplicando~a~propriedade~do~\underline{logaritmo~do~quociente}:\\\\\\\log\sqrt[5]{7,2}~=~\dfrac{1}{5}\cdot \left(\log72-\log10\right)\\\\\\\underline{Fatorando}~o~logaritmando~72:\\\\\\\log\sqrt[5]{7,2}~=~\dfrac{1}{5}\cdot \left(\log\,\left(2\cdot 2\cdot 2\cdot 3\cdot 3\right)-\log10\right)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}:\\\\\\\log\sqrt[5]{7,2}~=~\dfrac{1}{5}\cdot \left(\left(\log\log2+\log2+\log2+\log3+\log3\right)-\log10\right)$

$\sf Por~fim,~vamos~substituir~os~valores~dos~logaritmos~conhecidos:\\\\\\\log\sqrt[5]{7,2}~=~\dfrac{1}{5}\cdot \left(0,301+0,301+0,301+0,477+0,477~-~1\right)\\\\\\\log\sqrt[5]{7,2}~=~\dfrac{1}{5}\cdot \left(1,857~-~1\right)\\\\\\\log\sqrt[5]{7,2}~=~\dfrac{1}{5}\cdot \left(0,857\right)\\\\\\\boxed{\sf \log\sqrt[5]{7,2}~=~0,1714}$

b)

$\sf Agora,~reescrevendo~14,4~na~sua~\underline{forma~fracionaria}:\\\\\log14,4~=~\log\left(\dfrac{144}{10}\right)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~quociente}:\\\\\\\log14,4~=~\log144~-~\log10\\\\\\\underline{Fatorando}~o~logaritmando~144:\\\\\\\log14,4~=~\log\,(2\cdot 2\cdot 2\cdot 2\cdot 3\cdot 3)~-~\log10\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}:$

$\sf \log14,4~=~\left(\log2+\log2+\log2+\log2+\log3+\log3\right)~-~\log10$

$\sf Por~fim,~vamos~substituir~os~valores~dos~logaritmos~conhecidos:\\\\\\\log14,4~=~(0,301+0,301+0,301+0,301+0,477+0,477)-1\\\\\\\log14,4~=~2,158-1\\\\\\\boxed{\sf \log14,4~=~1,158}$

c)

$\sf \underline{Fatorando}~o~logaritmando~15:\\\\\\\log15~=~\log\,(3\cdot 5)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}:\\\\\\\log15~=~\log3+\log5\\\\\\\underline{Reescrevendo~o~logaritmando~5}~como~o~quociente~entre~10~e~2:\\\\\\\log15~=~\log3~+~\log\left(\dfrac{10}{2}\right)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~Quociente}:\\\\\\\log15~=~\log3~+~\left(\log10-\log2\right)\\\\\\\underline{Substituindo}~os~valores~dos~logaritmos~conhecidos:$

$\sf \log15~=~0,477+(1-0,301)\\\\\\\sf \log15~=~0,477+0,699\\\\\\\boxed{\sf\log15~=~1,176}$

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