Question

calcule o valor de x
3(x+1) = 3 × ㏒ 4 + 2 ㏒ √5 - ㏒ 32

Answer 1

Neste exercício utilizaremos algumas propriedades logaritmicas, assim convém lista-las previamente:

$\sf Propriedade~do~Logaritmo~da~Pot\hat{e}ncia:~\boxed{\sf \log_ba^c=c\cdot\log_ba}\\\\\\Propriedade~do~Logaritmo~do~Produto;~\boxed{\sf \log_b(a\cdot c)=\log_ba+\log_bc}$

Resolução

$\sf 3\cdot (x~+~1) ~=~ 3 \cdot \log4 ~+~ 2\cdot \log \sqrt{5} ~- ~\log 32\\\\\\Reescrever~4~e~32~como ~pot\hat{e}ncias~de~2\\\\\\\sf 3\cdot (x~+~1) ~=~ 3\cdot \log2^2 ~+~ 2\cdot \log \sqrt{5} ~- ~\log 2^5 \\\\\\Aplicar~a~propriedade~do~logaritmo~da~pot\hat{e}ncia\\\\\\\sf 3\cdot (x~+~1) ~=~ \log\left(2^2\right)^3 ~+~\log \left(\sqrt{5}\,\right)^2 ~+ ~\log \left(2^5\right)^{-1}\\\\\\\sf 3\cdot (x~+~1) ~=~ \log2^{2\cdot 3} ~+~\log 5 ~+~\log 2^{5\cdot (-1)}$

$\sf \sf 3\cdot (x~+~1) ~=~ \log2^6 ~+~\log 5 ~+ ~\log 2^{-5}\\\\\\\sf Reescrever~5~como~\dfrac{10}{2}\\\\\\\sf 3\cdot (x~+~1) ~=~ \log2^6 ~+~\log \dfrac{10}{2} ~+~\log 2^{-5}\\\\\\Aplicar~a~propriedade~do~logaritmo~do~produto\\\\\\3\cdot (x~+~1) ~=~ \log\left(2^6\cdot \dfrac{10}{2}\cdot 2^{-5}\right)\\\\\\3\cdot (x~+~1) ~=~ \log\left(10\cdot 2^{6+(-5)-1}\right)\\\\\\3\cdot (x~+~1) ~=~ \log\left(10\cdot 2^{0}\right)\\\\\\3\cdot (x~+~1) ~=~ \log\left(10\cdot 1\right)$

$\sf 3\cdot (x~+~1) ~=~ \log\left(10\cdot 1\right)\\\\\\3\cdot (x~+~1) ~=~ \log\left(10\right)\\\\\\3\cdot (x~+~1) ~=~ 1\\\\\\x+1~=~\dfrac{1}{3}\\\\\\x~=~\dfrac{1}{3}~-~1\\\\\\x~=~\dfrac{1~-~3}{3}\\\\\\\boxed{\sf x~=\,-\dfrac{2}{3}}$

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