Question

sendo log 2=a e log 3=b, calcule, em função de a e b.

a)
$log54$
b)
$log \sqrt[3]{12}$
c)
$log150$
d)
$log2.56$

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Answer 1

Neste exercício, vamos utilizar fatoração e algumas propriedades de logaritmos e de potências, que serão relembradas logo abaixo, para podermos reescrever os logaritmos dados em função de log2 e log3.

$\sf Propriedade~do~Logaritmo~do~Produto:~~\boxed{\sf \log_b(a\cdot c)=\log_ba+\log_bc}\\\\\\\sf Propriedade~do~Logaritmo~do~Quociente:~~\boxed{\sf \log_b\left(\dfrac{a}{c}\right)=\log_ba-\log_bc}\\\\\\\sf Propriedade~do~Logaritmo~da~Pot\hat{e}ncia:~~\boxed{\sf \log_ba^c=c\cdot \log_ba}\\\\\\Pot\hat{e}ncia~de~Expoente~Fracion\acute{a}rio:~~\boxed{\sf a^{^c/_b}~=~\sf \sqrt[b]{a^c}}$

a)

$\sf \underline{Fatorando}~o~logaritmando~54:\\\\\\\log54~=~\log\,(2\cdot 3\cdot 3\cdot 3)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}:\\\\\\\log54~=~\log2~+~\log3~+~\log3~+~\log3\\\\\\\underline{Substituindo}~os~valores~dos~logaritmos~conhecidos:\\\\\\\log54~=~a~+~b~+~b~+~b\\\\\\\boxed{\sf \log54~=~a~+~3b}$

b)

$\sf Transformando~a~raiz~c\acute{u}bica~numa~\underline{pot\hat{e}ncia~de~expoente~fracion\acute{a}rio}:\\\\\\\log\sqrt[3]{12}~=~\log12^{^1/_3}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~da~pot\hat{e}ncia}:\\\\\\\log\sqrt[3]{12}~=~\dfrac{1}{3}\cdot \log12\\\\\\\underline{Fatorando}~o~logaritmando~12:\\\\\\\log\sqrt[3]{12}~=~\dfrac{1}{3}\cdot \log\,(2\cdot 2\cdot 3)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}:$

$\sf \log\sqrt[3]{12}~=~\dfrac{1}{3}\cdot\left(\log2~+~\log2~+~\log3\right)\\\\\\\underline{Substituindo}~os~valores~dos~logaritmos~conhecidos:\\\\\\\log\sqrt[3]{12}~=~\dfrac{1}{3}\cdot\left(a~+~a~+~b\right)\\\\\\\boxed{\sf \log\sqrt[3]{12}~=~\dfrac{2a~+~b}{3}}$

c)

$\sf \underline{Fatorando}~o~logaritmando~150:\\\\\\\log150~=~\log\,(2\cdot 3\cdot 5\cdot 5)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}:\\\\\\\log150~=~\log2~+~\log3~+~\log5~+~\log5\\\\\\Note~que~n\tilde{a}o~temos~o~valor~de~log5,~mas~podemos~contornar~este\\problema~reescrevendo~5~como~^{10}/_2:\\\\\\\log150~=~\log2~+~\log3~+~\log\left(\dfrac{10}{2}\right)~+~\log\left(\dfrac{10}{2}\right)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~quociente}:$

$\sf \log150~=~\log2~+~\log3~+~\left(\log10-\log2\right)~+~\left(\log10-\log2\right)\\\\\\\underline{Substituindo}~os~valores~dos~logaritmos~conhecidos:\\\\\\\log150~=~a~+~b~+~(1-a)~+~(1-a)\\\\\\\log150~=~a~+~b~+~1~+~1~-~a~-~a\\\\\\\boxed{\sf \log150~=~2~-~a~+~b}$

d)

$\sf Reescrevendo~2,56~na~sua ~\underline{forma~fracion\acute{a}ria}:\\\\\\\log2,56~=~\log\left(\dfrac{256}{100}\right)\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~quociente}:\\\\\\\log2,56~=~\log256~-~\log100\\\\\\\sf \underline{Fatorando}~os~logaritmandos~256~e~100:\\\\\\\log2,56~=~\log\,(2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2\cdot 2)~-~\log\,(10\cdot 10)\\\\\\\log2,56~=~\log2^8~-~\log10^2$

$\sf Aplicando~a~propriedade~do~\underline{logaritmo~da~pot\hat{e}ncia}:\\\\\\\log2,56~=~8\cdot \log2~-~2\cdot \log10\\\\\\\underline{Substituindo}~os~valores~dos~logaritmos~conhecidos:\\\\\\\log2,56~=~8\cdot a~-~2\cdot 1\\\\\\\boxed{\sf \log2,56~=~8a-2}$

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