Question

Usando os valores aproximados de log 2 ≅ 0,30, log 3 ≅ 0,47 e log 7 ≅ 0,85. Calcule: a) log 14 84 b) log21 5 c) log6 15

Answer 1

Vamos aplicar as propriedades dos logaritmos para reescrever as expressões de tal forma que possamos utilizar as informações dadas no enunciado para calcula-las.

a)

$\log_{_{14}}84~=\\\\\\Aplicando~a~propriedade~da~\underline{troca~de~base}\\\\\\=~\dfrac{\log84}{\log14}\\\\\\\underline{Fatorando}~os~logaritmandos\\\\\\=~\dfrac{\log\,(2\cdot2\cdot3\cdot7)}{\log\,(2\cdot7)}\\\\\\Aplicando~a~propriedade~do~\underline{logaritmo~do~produto}\\\\\\=~\dfrac{\log2~+~\log2~+~\log3~+~\log7}{\log2~+~\log7}\\\\\\=~\dfrac{0,30~+~0,30~+~0,47~+~0,85}{0,30~+~0,85}\\\\\\=~\dfrac{1,92}{1,15}\\\\\\=~\boxed{\dfrac{192}{115}}\\\\ou\\\\\approx~\boxed{1,67}$

b)

$\log_{_{21}}5~=\\\\\\Aplicando~a~propriedade~da~\underline{troca~de~base}\\\\\\=~\dfrac{\log5}{\log21}\\\\\\\underline{Fatorando}~o~logaritmando~21~e~reecrevendo~o~logaritmando\\5~como~10/2\\\\\\=~\dfrac{\log\,\left(\dfrac{10}{2}\right)}{\log\,(3\cdot7)}\\\\\\Aplicando~as~propriedades~do~\underline{logaritmo~do~produto}~e~do\\\underline{logaritmo~do~quociente}\\\\\\=~\dfrac{\log10~-~\log2}{\log3~+~\log7}$

$=~\dfrac{1~-~0,30}{0,47~+~0,85}\\\\\\=~\dfrac{0,70}{1,32}\\\\\\=~\boxed{\dfrac{35}{66}}\\\\ou\\\\\approx~\boxed{0,53}$

c)

$\log_{_{6}}15~=\\\\\\Aplicando~a~propriedade~da~\underline{troca~de~base}\\\\\\=~\dfrac{\log15}{\log6}\\\\\\\underline{Fatorando}~os~logaritmandos\\\\\\=~\dfrac{\log\,(3\cdot5)}{\log\,(2\cdot3)}\\\\\\Reescrevendo~5~como~10/2\\\\\\=~\dfrac{\log\,\left(3\cdot\dfrac{10}{2}\right)}{\log\,(2\cdot3)}\\\\\\Aplicando~as~propriedades~do~\underline{logaritmo~do~produto}~e~do\\\underline{logaritmo~do~quociente}\\\\\\=~\dfrac{\log3~+~\log\,\left(\dfrac{10}{2}\right)}{\log2~+~\log3}$

$=~\dfrac{\log3~+~\left(\log10~-~\log2\right)}{\log2~+~\log3}\\\\\\=~\dfrac{0,47~+~1~-~0,30}{0,30~+~0,47}\\\\\\=~\dfrac{1,17}{0,77}\\\\\\=~\boxed{\dfrac{117}{77}}\\\\ou\\\\\approx~\boxed{1,52}$