Question

Resolva no campo dos complexos, a equação..

$\Large\boxed{\log_{(0,2)}(625x)+\log_x( \sqrt{5})=1-\log_{125}(0,0016x)}$

Answer 1

$\mathrm{\ell og_{\left(0,2 \right )}}\left(625x \right )+\mathrm{\ell og}_{x}\left(\sqrt{5} \right )=1-\mathrm{\ell og_{125}}\left(0,0016x \right )$


Antes de tudo, vamos converter os logaritmos para uma base comum utilizando a lei de mudança de base. Vamos mudar todos os logaritmos para a base $5$:

$\dfrac{\mathrm{\ell og_{5}}\left(625x \right )}{\mathrm{\ell og_{5}}\left(0,2 \right )}+\dfrac{\mathrm{\ell og_{5}}{\left(\sqrt{5} \right )}}{\mathrm{\ell og_{5}\,}x}=1-\dfrac{\mathrm{\ell og_{5}}{\left(0,0016x \right )}}{\mathrm{\ell og_{5}\,}125}$

$\dfrac{\mathrm{\ell og_{5}\,}625+\mathrm{\ell og_{5}\,}x}{\mathrm{\ell og_{5}}\left(5^{-1} \right )}+\dfrac{\mathrm{\ell og_{5}}{\left(5^{1/2} \right )}}{\mathrm{\ell og_{5}\,}x}=1-\dfrac{\mathrm{\ell og_{5}}{\left(0,0016 \right )}+\mathrm{\ell og_{5}\,}x}{\mathrm{\ell og_{5}}\left(5^{3} \right )}\\ \\ \dfrac{\mathrm{\ell og_{5}}{\left(5^{4} \right )}+\mathrm{\ell og_{5}\,}x}{-1}+\dfrac{1/2}{\mathrm{\ell og_{5}\,}x}=1-\dfrac{\mathrm{\ell og_{5}}{\left(5^{-4} \right )}+\mathrm{\ell og_{5}\,}x}{3}\\ \\ \dfrac{4+\mathrm{\ell og_{5}\,}x}{-1}+\dfrac{1}{2\,\mathrm{\ell og_{5}\,}x}=1-\dfrac{-4+\mathrm{\ell og_{5}\,}x}{3}$


Para evitar carrregar a notação de logaritmos até o final da resolução, vamos fazer uma mudança de variável:

$y=\mathrm{\ell og_{5}\,}x$


Substituindo na equação, temos

$\dfrac{4+y}{-1}+\dfrac{1}{2y}=1-\dfrac{-4+y}{3}\\ \\ \dfrac{-2y\left(4+y \right )+1}{2y}=\dfrac{3-\left(-4+y \right )}{3}\\ \\ \dfrac{-8y-2y^{2}+1}{2y}=\dfrac{3+4-y}{3}\\ \\ \dfrac{-8y-2y^{2}+1}{2y}=\dfrac{7-y}{3}\\ \\ 3\left(-8y-2y^{2}+1 \right )=2y\left(7-y \right )\\ \\ -24y-6y^{2}+3=14y-2y^{2}\\ \\ -2y^{2}+6y^{2}+14y+24y-3=0\\ \\ 4y^{2}+38y-3=0\;\;\Rightarrow\;\;a=4,\;b=38,\;c=-3$


$\Delta=b^{2}-4ac\\ \\ \Delta=38^{2}-4\cdot 4\cdot \left(-3 \right )\\ \\ \Delta=1444+48\\ \\ \Delta=1492\\ \\ \Delta=2^{2}\cdot 373 \\ \\ \\ y=\dfrac{-b\pm \sqrt{\Delta}}{2a}\\ \\ y=\dfrac{-38\pm \sqrt{2^{2}\cdot 373}}{2\cdot 4}\\ \\ y=\dfrac{-38\pm 2\sqrt{373}}{2\cdot 4}\\ \\ y=\dfrac{\diagup\!\!\!\! 2\cdot\left(-19 \pm \sqrt{373} \right )}{\diagup\!\!\!\! 2\cdot 4}\\ \\ y=\dfrac{-19 \pm \sqrt{373}}{4}$

$\begin{array}{rcl} y=\dfrac{-19 + \sqrt{373}}{4}&\text{ ou }&y=\dfrac{-19 - \sqrt{373}}{4} \end{array}$


Voltando à variável original $x$, temos

$\begin{array}{rcl} \mathrm{\ell og_{5}\,}x=\dfrac{-19 + \sqrt{373}}{4}&\text{ ou }&\mathrm{\ell og_{5}\,}x=\dfrac{-19 - \sqrt{373}}{4} \end{array}\\ \\ \\ \boxed{ \begin{array}{rcl} x=5^{\frac{-19 + \sqrt{373}}{4}}&\text{ ou }&x=5^{\frac{-19 - \sqrt{373}}{4}} \end{array} }$

Answer 2

Explicação passo-a-passo:

$log_{0,2}(625x) + log_{x}( \sqrt{5} ) = 1 - log_{125}(0,0016x) \\ log_{0,2}(625x) + log_{x}( \sqrt{5} ) = 1 - log_{125}(0,0016x), \: x \: E \: (0 ,1) \: U \: (1, + \infty ) \\ log_{ {5}^{ - 1} }(625x) + log_{ x }( \sqrt{5} ) = 1 - log_{ {5}^{3} }(0,0016x) \\ - log_{5}(625x) + log_{x}( \sqrt{5} ) = 1 - \frac{1}{3} \times log_{5}(0,0016x) \\ - ( log_{5}(625) + log_{5}(x) ) + log_{x}( \sqrt{5} ) = 1 - \frac{1}{3} \times ( log_{5}(0,0016) + log_{5}(x) ) \\ - ( log_{5}( {5}^{4} ) + log_{5}(x) ) + log_{x}( \sqrt{5} ) = 1 - \frac{1}{3} \times ( log_{5}( {5}^{ - 4} ) + log_{5}(x) ) \\ - (4 + log_{5}(x) ) + log_{ x}( \sqrt{5} ) = 1 - \frac{1}{3} \times ( - 4 + log_{5}(x) ) \\ - 4 - log_{5}(x) + log_{x}( \sqrt{5} ) = 1 + \frac{4}{3} - \frac{1}{3} \times log_{5}(x) \\ - 4 - log_{5}(x) + log_{x}( \sqrt{5} ) = \frac{7}{3} - \frac{1}{3} \times log_{5}(x) \\ - 12 - 3 log_{5}(x) + 3 log_{x}( \sqrt{5} ) = 7 - log_{5}(x) \\ - 3 log_{5}(x) + 3 log_{x}( \sqrt{5} ) + log_{5}(x) =7 + 12 \\ - 2 log_{5}(x) + 3 \times \frac{ log_{5}( \sqrt{5} ) }{ log_{5}(x) } = 19 \\ - 2 log_{5}(x) + 3 \times \frac{ log_{5}( {5}^{ \frac{1}{2} } ) }{ log_{5}(x) } = 19 \\ - 2 log_{5}(x) + 3 \times \frac{ \frac{1}{2} }{ log_{5}(x) } = 19 \\ - 2 log_{5}(x) + 3 \times \frac{1}{2 log_{5}(x) } = 19 \\ - 2 log_{5}(x) + \frac{3}{2 log_{5}(x) } = 19 \\ - 2t + \frac{3}{2t} = 19 \\ \\ t = \frac{ - 19 + \sqrt{373} }{4} \\ t = \frac{ - 19 - \sqrt{373} }{4} \\ \\ log_{5}(x) = \frac{ - 19 + \sqrt{373} }{4} \\ log_{5}(x) = \frac{ - 19 - \sqrt{373} }{4} \\ \\ x = {5}^{\frac{ - 19 + \sqrt{373} }{4} } \\ x = {5}^{\frac{ - 19 - \sqrt{373} }{4}} , \: x \: \: e \: \: (0,1)u(1, + \infty ) \\ \\ x _{1} = {5}^{ \frac{ - 19 + \sqrt{373} }{4} } ,x _{2} = {5}^{ \frac{ - 19 - \sqrt{373} }{4} }$

• Espero ter ajudado.