Respostas
respondido por:
2
Espero ter ajudado
respondido por:
3
Olá Jenifer,
no logaritmo
, vamos decompor o 108 em fatores
primos, então o logaritmo se transformará em:
108|2
54|2
27|3
9|3
3|3______
1| 2² · 3³
![\log( \sqrt{108})=\log( \sqrt{2^2\cdot3^3} )\\\log(\sqrt{108})=\log({2^2\cdot3^3})^{\tfrac{1}{2} \log( \sqrt{108})=\log( \sqrt{2^2\cdot3^3} )\\\log(\sqrt{108})=\log({2^2\cdot3^3})^{\tfrac{1}{2}](https://tex.z-dn.net/?f=%5Clog%28+%5Csqrt%7B108%7D%29%3D%5Clog%28+%5Csqrt%7B2%5E2%5Ccdot3%5E3%7D+%29%5C%5C%5Clog%28%5Csqrt%7B108%7D%29%3D%5Clog%28%7B2%5E2%5Ccdot3%5E3%7D%29%5E%7B%5Ctfrac%7B1%7D%7B2%7D)
Agora, aplicando a 3a propriedade de log, a da potência,
![\large\boxed{\log(b)^n=n\cdot\log(b)} \large\boxed{\log(b)^n=n\cdot\log(b)}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Clog%28b%29%5En%3Dn%5Ccdot%5Clog%28b%29%7D)
![\log( \sqrt{108})= \dfrac{1}{2}\cdot \log( \sqrt{2^2\cdot3^3}) \log( \sqrt{108})= \dfrac{1}{2}\cdot \log( \sqrt{2^2\cdot3^3})](https://tex.z-dn.net/?f=%5Clog%28+%5Csqrt%7B108%7D%29%3D+%5Cdfrac%7B1%7D%7B2%7D%5Ccdot+%5Clog%28+%5Csqrt%7B2%5E2%5Ccdot3%5E3%7D%29)
Podemos também, aplicar a primeira propriedade de log, a do produto:
![\large\boxed{\log(a\cdot b)=\log(a)+\log(b)} \large\boxed{\log(a\cdot b)=\log(a)+\log(b)}](https://tex.z-dn.net/?f=%5Clarge%5Cboxed%7B%5Clog%28a%5Ccdot+b%29%3D%5Clog%28a%29%2B%5Clog%28b%29%7D)
![\log( \sqrt{108})= \dfrac{1}{2}\cdot[\log(2^2)+\log(3^3)] \log( \sqrt{108})= \dfrac{1}{2}\cdot[\log(2^2)+\log(3^3)]](https://tex.z-dn.net/?f=%5Clog%28+%5Csqrt%7B108%7D%29%3D+%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%5B%5Clog%282%5E2%29%2B%5Clog%283%5E3%29%5D+)
E novamente a da potência:
![\log( \sqrt{108})= \dfrac{1}{2}\cdot[2\cdot \log(2)+3\cdot\log(3)] \log( \sqrt{108})= \dfrac{1}{2}\cdot[2\cdot \log(2)+3\cdot\log(3)]](https://tex.z-dn.net/?f=%5Clog%28+%5Csqrt%7B108%7D%29%3D+%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%5B2%5Ccdot+%5Clog%282%29%2B3%5Ccdot%5Clog%283%29%5D)
Substitua então os valores de log dados, log2=0,3 e log3=0,47:
![\log( \sqrt{108})= \dfrac{1}{2}\cdot[2\cdot0,3+3\cdot0,47]\\\\
\log( \sqrt{108})= \dfrac{1}{2}\cdot[0,6+1,41]\\\\
\log( \sqrt{108})= \dfrac{1}{2}\cdot2,01\\\\
\log( \sqrt{108})= \dfrac{2,01}{2}\\\\
\Large\boxed{\boxed{\log( \sqrt{108})=1,005}} \log( \sqrt{108})= \dfrac{1}{2}\cdot[2\cdot0,3+3\cdot0,47]\\\\
\log( \sqrt{108})= \dfrac{1}{2}\cdot[0,6+1,41]\\\\
\log( \sqrt{108})= \dfrac{1}{2}\cdot2,01\\\\
\log( \sqrt{108})= \dfrac{2,01}{2}\\\\
\Large\boxed{\boxed{\log( \sqrt{108})=1,005}}](https://tex.z-dn.net/?f=%5Clog%28+%5Csqrt%7B108%7D%29%3D+%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%5B2%5Ccdot0%2C3%2B3%5Ccdot0%2C47%5D%5C%5C%5C%5C%0A%5Clog%28+%5Csqrt%7B108%7D%29%3D+%5Cdfrac%7B1%7D%7B2%7D%5Ccdot%5B0%2C6%2B1%2C41%5D%5C%5C%5C%5C%0A%5Clog%28+%5Csqrt%7B108%7D%29%3D+%5Cdfrac%7B1%7D%7B2%7D%5Ccdot2%2C01%5C%5C%5C%5C%0A%5Clog%28+%5Csqrt%7B108%7D%29%3D+%5Cdfrac%7B2%2C01%7D%7B2%7D%5C%5C%5C%5C%0A%5CLarge%5Cboxed%7B%5Cboxed%7B%5Clog%28+%5Csqrt%7B108%7D%29%3D1%2C005%7D%7D)
Tenha ótimos estudos ;P
no logaritmo
primos, então o logaritmo se transformará em:
108|2
54|2
27|3
9|3
3|3______
1| 2² · 3³
Agora, aplicando a 3a propriedade de log, a da potência,
Podemos também, aplicar a primeira propriedade de log, a do produto:
E novamente a da potência:
Substitua então os valores de log dados, log2=0,3 e log3=0,47:
Tenha ótimos estudos ;P
Perguntas similares
7 anos atrás
7 anos atrás
9 anos atrás
9 anos atrás
9 anos atrás