alguem pode me ajudar?
1/7= raiz setima de 49^x-1
Soluções para a tarefa
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Eleva os dois lados a 7, o que vai acabar c a raiz, de um lado vai ficar 1/49 que é igual a 49^-1, o que torna as bases iguais podendo igualar os expoentes e descobrir o valor de x
Respondido por
6
Olá,
use as propriedades da exponenciação..
![\dfrac{1}{7}= \sqrt[7]{49^{x-1}}\\
\sqrt[7]{(7^2)^{x-1}}= \dfrac{1}{7^1}\\\\
\sqrt[7]{7^{2x-2}}=7^{-1}\\\\
\not7^{ \tfrac{2x-2}{7} }=\not7^{-1}\\\\
\dfrac{2x-2}{7}=-1\\\\
2x-2=(-1)\cdot7\\
2x-2=-7\\
2x=-7+2\\
2x=-5\\\\
x=- \dfrac{5}{2}\\\\\\
\Large\boxed{S=\left\{- \dfrac{5}{2}\right\}}](https://tex.z-dn.net/?f=+%5Cdfrac%7B1%7D%7B7%7D%3D+%5Csqrt%5B7%5D%7B49%5E%7Bx-1%7D%7D%5C%5C%0A+%5Csqrt%5B7%5D%7B%287%5E2%29%5E%7Bx-1%7D%7D%3D+%5Cdfrac%7B1%7D%7B7%5E1%7D%5C%5C%5C%5C%0A+%5Csqrt%5B7%5D%7B7%5E%7B2x-2%7D%7D%3D7%5E%7B-1%7D%5C%5C%5C%5C%0A%5Cnot7%5E%7B+%5Ctfrac%7B2x-2%7D%7B7%7D+%7D%3D%5Cnot7%5E%7B-1%7D%5C%5C%5C%5C%0A+%5Cdfrac%7B2x-2%7D%7B7%7D%3D-1%5C%5C%5C%5C%0A2x-2%3D%28-1%29%5Ccdot7%5C%5C%0A2x-2%3D-7%5C%5C%0A2x%3D-7%2B2%5C%5C%0A2x%3D-5%5C%5C%5C%5C%0Ax%3D-+%5Cdfrac%7B5%7D%7B2%7D%5C%5C%5C%5C%5C%5C%0A%5CLarge%5Cboxed%7BS%3D%5Cleft%5C%7B-+%5Cdfrac%7B5%7D%7B2%7D%5Cright%5C%7D%7D++++++++ "\dfrac{1}{7}= \sqrt[7]{49^{x-1}}\\
\sqrt[7]{(7^2)^{x-1}}= \dfrac{1}{7^1}\\\\
\sqrt[7]{7^{2x-2}}=7^{-1}\\\\
\not7^{ \tfrac{2x-2}{7} }=\not7^{-1}\\\\
\dfrac{2x-2}{7}=-1\\\\
2x-2=(-1)\cdot7\\
2x-2=-7\\
2x=-7+2\\
2x=-5\\\\
x=- \dfrac{5}{2}\\\\\\
\Large\boxed{S=\left\{- \dfrac{5}{2}\right\}}")
Tenha ótimos estudos ;D
use as propriedades da exponenciação..
Tenha ótimos estudos ;D
ennygontier:
obrigadaa
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