Question

Dado que log de 2 na base 14 = r e log de 3 na base 14 = s , calcule:
a)log de 7 na base 14 
b) log 8 na base 14 
c) log de √1,5 na base 14 
d) log de 72 na base 14
e) log de 2 na base 3

Answer 1

Neste exercício, vamos utilizar as seguintes propriedades de logaritmo e de potência:

$(1)~Logaritmo~do~Produto:~~\log_{_b}(a\cdot c)~=~\log_{_b}a~+~\log_{_b}c\\\\(2)~Logaritmo~do~Quociente:~~\log_{_b}\left(\dfrac{a}{c}\right)~=~\log_{_b}a~-~\log_{_b}c\\\\(3)~Logaritmo~da~Potencia:~~\log_{_b}a^{\frac{c}{d}}~=~\dfrac{c}{d}\cdot\log_{_b}a\\\\(4)~Troca~de~Bases:~~\log_{_b}a~=~\dfrac{\log_{_c}a}{\log_{_c}b}\\\\(5)~Potencia~de~Expoente~Fracionario:~~a^{\frac{b}{c}}~=~\sqrt[c]{a^b}$

a)

$\log_{_{14}}7~=\\\\\\Podemos~reescrever~o~logaritmando~como~\dfrac{14}{2}\\\\\\=~\log_{_{14}}\left(\dfrac{14}{2}\right)\\\\\\Utilizando~a~propriedade~do~logaritmo~do~quociente:\\\\\\=~\log_{_{14}}14~-~\log_{_{14}}2\\\\\\Substituindo~os~valores~dos~logaritmos:\\\\\\=~\boxed{1~-~r}~~\Rightarrow~Resposta$

b)

$\log_{_{14}}8~=\\\\\\Fatorando~o~logaritmando:\\\\\\=~\log_{_{14}}\,(2\cdot 2\cdot 2)\\\\\\Utilizando~a~propriedade~do~logaritmo~do~produto:\\\\\\=~\log_{_{14}}2~+~\log_{_{14}}2~+~\log_{_{14}}2\\\\\\Substituindo~os~valores~dos~logaritmos:\\\\\\=~r~+~r~+~r\\\\\\=~\boxed{3r}~~ \Rightarrow~Resposta$

c)

$\log_{_{14}}\sqrt{1,5}~=\\\\\\Reescrevendo~1,5~na~sua~forma~fracionaria~reduzida:\\\\\\=~\log_{_{14}}\sqrt{\dfrac{3}{2}}\\\\\\Utilizando~a~propriedade~da~potencia~de~expoente~fracionario:\\\\\\=~\log_{_{14}}\left(\dfrac{3}{2}\right)^{\frac{1}{2}}\\\\\\Utilizando~a~propriedade~do~logaritmo~da~potencia:\\\\\\=~\dfrac{1}{2}\cdot \log_{_{14}}\left(\dfrac{3}{2}\right)\\\\\\Utilizando~a~propriedade~do~logaritmo~do~quociente:$

$=~\dfrac{1}{2}\cdot \left(\log_{_{14}}3~-~\log_{_{14}}2\right)\\\\\\Substituindo~os~valores~dos~logaritmos:\\\\\\=~\boxed{\dfrac{1}{2}\cdot \left(r~-~s\right)}~~ou~~ \boxed{\dfrac{r~-~s}{2}}~~\Rightarrow~Resposta$

d)

$\log_{_{14}}72~=\\\\\\Fatorando~o~logaritmando:\\\\\\=~\log_{_{14}}\,(2\cdot 2\cdot 2\cdot 3\cdot 3)\\\\\\Utilizando~a~propriedade~do~logaritmo~do~produto:\\\\\\=~\log_{_{14}}2~+~\log_{_{14}}2~+~\log_{_{14}}2~+~\log_{_{14}}3~+~\log_{_{14}}3\\\\\\Substituindo~os~valores~dos~logaritmos:\\\\\\=~r~+~r~+~r~+~s~+~s\\\\\\=~\boxed{3r~+~2s}~~\Rightarrow~Resposta$

e)

$\log_{_{3}}2~=\\\\\\Utilizando~a~propriedade~da~troca~de~base:\\\\\\=~\dfrac{\log_{_{14}}2}{\log_{_{14}}3}\\\\\\Substituindo~os~valores~dos~logaritmos:\\\\\\=~\boxed{~\dfrac{r}{s}~}~~\Rightarrow~Resposta\\\\\\\\\Huge{\begin{array}{c}\Delta \tt{\!\!\!\!\!\!\,\,o}\!\!\!\!\!\!\!\!\:\,\perp\end{array}}Qualquer~d\acute{u}vida,~deixe~ um~coment\acute{a}rio$