Matemática, perguntado por manusbb, 1 ano atrás

sendo log de a na base 3= r, log de b na base 3 = s, e log de c na base 3 = t, obtenha log3 a2 .raiz de b/raiz cubica de c, em funçao de r,s,t.

Soluções para a tarefa

Respondido por Niiya
3
Propriedades usadas:

\boxed{\boxed{\sqrt[n]{a^{m}}=a^{m/n}}}\\\\\\\boxed{\boxed{log_{b}(x\cdot y)=log_{b}(x)+log_{b}(y)}}\\\\\\\boxed{\boxed{log_{b}\left(\dfrac{x}{y}\right)=log_{b}(x)-log_{b}(y)}}\\\\\\\boxed{\boxed{log_{b}(a^{n})=n\cdot log_{b}(a)}}
__________________________________

log_{3}\left(\dfrac{a^{2}\sqrt{b}}{\sqrt[3]{c}}\right)=log_{3}(a^{2})+log_{3}(\sqrt{b})-log_{3}\sqrt[3]{c}\\\\\\log_{3}\left(\dfrac{a^{2}\sqrt{b}}{\sqrt[3]{c}}\right)=2\cdot log_{3}(a)+log_{3}(b^{1/2})-log_{3}(c^{1/3})\\\\\\log_{3}\left(\dfrac{a^{2}\sqrt{b}}{\sqrt[3]{c}}\right)=2\cdot log_{3}(a)+\dfrac{1}{2}log_{3}(b)-\dfrac{1}{3}log_{3}(c)

Sustituindo os logaritmos por r, s e t:

log_{3}\left(\dfrac{a^{2}\sqrt{b}}{\sqrt[3]{c}}\right)=2\cdot r+\dfrac{1}{2}\cdot s-\dfrac{1}{3}\cdot t\\\\\\log_{3}\left(\dfrac{a^{2}\sqrt{b}}{\sqrt[3]{c}}\right)=\dfrac{12r}{6}+\dfrac{3s}{6}-\dfrac{2t}{6}\\\\\\\boxed{\boxed{log_{3}\left(\dfrac{a^{2}\sqrt{b}}{\sqrt[3]{c}}\right)=\dfrac{1}{6}(12r+3s-2t)}}
Perguntas interessantes