Question

•7logx - logx/4 =13 + 1/logx
Alguém me ajuda nessa questão? Queria saber como chegar na resposta...

Answer 1

$7\cdot\text{log}~x-\dfrac{\text{log}~x}{4}=13+\dfrac{1}{\text{log}~x}$

Seja $\text{log}~x=y$

$7y-\dfrac{y}{4}=13+\dfrac{1}{y}$

$\text{mmc}(4,y)=4y$

$28y^2-y^2=52y+4$

$27y^2-52y-4=0$

$\Delta=(-52)^2-4\cdot27\cdot(-4)=2704+432=3136$

$y=\dfrac{-(-52)\pm\sqrt{3136}}{2\cdot27}=\dfrac{52\pm56}{54}$

$y'=\dfrac{52+56}{54}~\longrightarrow~y'=\dfrac{108}{54}~\longrightarrow~y'=2$

$y"=\dfrac{52-56}{54}~\longrightarrow~y"=\dfrac{-4}{54}~\longrightarrow~y"=-\dfrac{2}{27}$

Logo:

$\text{log}~x=2$ ou $\text{log}~x=-\dfrac{2}{27}$

$x'=10^2~\longrightarrow~x'=100$

$x"=10^{-\frac{2}{27}}~\longrightarrow~x"=\sqrt[27]{10^{-2}}~\longrightarrow~x"=\sqrt[27]{\dfrac{1}{100}}$

ou $x"=\dfrac{1}{10^{\frac{2}{27}}}$

$\text{S}=\{100,\frac{1}{10^{\frac{2}{27}}}\}$