Question

Questao 2 Sabendo que log2=0,30 e log3=0,48, calcule cada um dos logaritmos a seguir. a) log16 b) log 0,2 c) log6 d) log18 e) log 5

Answer 1

Para resolver sua questão, devemos saber as propriedades dos logs. As que iremos utilizar na resolução são as seguintes:

$\Large\displaystyle\begin{gathered}\sf \log(a\cdot b)= \log(a)+\log(b) \end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log\left(\frac{a}{b} \right)= \log(a)-\log(b) \end{gathered}$

Sabendo que $\large\displaystyle\begin{gathered}\sf \log(2)=0,30 \end{gathered}$ e $\large\displaystyle\begin{gathered}\sf \log(3)=0,48 \end{gathered}$, temos que:

$\Large\displaystyle\begin{gathered}\sf \log(16)=\log(2^4) \end{gathered}$

E pela propriedade do "peteleco" ( $\large\displaystyle\begin{gathered}\sf \log(a^n)=n\cdot \log(a) \end{gathered}$ ), ficamos da seguinte forma:

$\Large\displaystyle\begin{gathered}\sf \log(16)=4\log(2) \end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(16)=4\cdot 0,3 \end{gathered}$

$\Large\displaystyle\begin{gathered}\boxed{\sf \log(16)=1,2} \end{gathered}$

Já na letra b, temos que utilizar a segunda propriedade.

$\Large\displaystyle\begin{gathered}\sf \log(0,2)=\log\left(\frac{2}{10} \right) \end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(0,2)=\log(2)-\log(10) \end{gathered}$

Lembrando que, o logaritmo de mesma base é sempre igual a um.

$\Large\displaystyle\begin{gathered}\sf \log(0,2)=0,3-1 \end{gathered}$

$\Large\displaystyle\begin{gathered}\boxed{\sf \log(0,2)=-0,7}\end{gathered}$

Na letra c, utilizaremos a primeira propriedade.

$\Large\displaystyle\begin{gathered}\sf \log(6)=\log(2\cdot 3)\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(6)=\log(2)+\log(3)\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(6)=0,3+0,48\end{gathered}$

$\Large\displaystyle\begin{gathered}\boxed{\sf \log(6)=0,78}\end{gathered}$

Na letra d, teremos que fatorar o número 18.

$\Large\displaystyle\begin{gathered}\sf \log(18)=\log(2\cdot 3^2)\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(18)=\log(2)+\log(3^2)\end{gathered}$

E pela propriedade do peteleco:

$\Large\displaystyle\begin{gathered}\sf \log(18)=\log(2)+2\log(3)\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(18)=0,3+0,96\end{gathered}$

$\Large\displaystyle\begin{gathered}\boxed{\sf \log(18)=1,26}\end{gathered}$

E por fim, na letra e, temos que:

$\Large\displaystyle\begin{gathered}\sf \log(5)=\log(2,5\cdot 2)\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(5)=\log\left( \frac{25}{10} \right)+\log(2)\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(5)=\log( 25)-\log(10)+\log(2)\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(5)=\log( 5^2)-1+0,3\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(5)=2\log( 5)-0,7\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf \log(5)-2\log( 5)=-0,7\end{gathered}$

$\Large\displaystyle\begin{gathered}\sf -\log( 5)=-0,7\end{gathered}$

$\Large\displaystyle\begin{gathered}\boxed{\sf \log( 5)=0,7}\end{gathered}$

Qualquer dúvida quanto a resolução dada é só chamar!

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