Question

Resolva a equação logarítmica

$\left( \dfrac{2-log_4x}{1+log_4x}\right)^2=2+log_2\left( \dfrac{1}{x}\right)$

Answer 1

Observe que:

$log_2\left(\frac{1}{x}\right)=log_2x^{-1}=-log_2x=-\frac{log_4x}{log_42}=-\frac{log_4x}{\frac{1}{2}}=-2log_4x$

Substituindo na equação original:

$\left(\frac{2-log_4x}{1+log_4x}\right)^2=2-2log_4x\\ \\ Substituindo: \ \ y=log_4x\\ \\ \left(\frac{2-y}{1+y}\right)^2=2-2y\\ \\ \frac{4-4y+y^2}{1+2y+y^2}=2-2y\\ \\ (2-2y)(1+2y+y^2)=4-4y+y^2\\ \\ -2 y^3-2 y^2+2 y+2=4-4y+y^2\\ \\ -2 y^3-2 y^2+2 y+2-4+4y-y^2=0\\ \\ -2y^3-3y^2+6y-2=0$

Pesquisando as raízes desta equação (por qualquer método), temos:

$S=\{\frac{1}{2}; -1-\sqrt3; \sqrt3-1 \}$

Substituindo:

$log_4x=\frac{1}{2}\\ \\ \boxed{x_1=4^{\frac{1}{2}}=2}\\ \\ log_4x=-1-\sqrt3\\ \\ \boxed{x_2=4^{-1-\sqrt3}}\\ \\ log_4x=\sqrt3-1\\ \\ \boxed{x_3=4^{\sqrt3-1}}$

Answer 2

Explicação passo-a-passo:

$\sf \left(\dfrac{2-log_{4}~x}{1+log_{4}~x}\right)^2=2+log_{2}~\dfrac{1}{x}$

$\sf \left(\dfrac{2-\frac{log_{2}~x}{log_{2}~4}}{1+\frac{log_{2}~x}{log_{2}~4}}\right)^2=2+log_{2}~x^{-1}$

$\sf \left(\dfrac{2-\frac{log_{2}~x}{2}}{1+\frac{log_{2}~x}{2}}\right)^2=2+(-1)\cdot log_{2}~x$

$\sf \left(\dfrac{\frac{4-log_{2}~x}{2}}{\frac{2+log_{2}~x}{2}}\right)^2=2-log_{2}~x$

$\sf \left(\dfrac{4-log_{2}~x}{2+log_{2}~x}\right^2=2-log_{2}~x$

Seja $\sf k=log_{2}~x$

$\sf \left(\dfrac{4-k}{2+k}\right)^2=2-k$

$\sf \dfrac{16-8k+k^2}{4+4k+k^2}=2-k$

$\sf k^2-8k+16=(4+4k+k^2)\cdot(2-k)$

$\sf k^2-8k+16=8+8k+2k^2-4k-4k^2-k^3$

$\sf k^2-8k+16=-k^3-2k^2+4k+8$

$\sf k^3+k^2+2k^2-8k-4k+16-8=0$

$\sf k^3+3k^2-12k+8=0$

$\sf k^3-k^2+4k^2-4k-8k+8=0$

$\sf k^2\cdot(k-1)+4k\cdot(k-1)-8\cdot(k-1)=0$

$\sf (k-1)\cdot(k^2+4k-8)=0$

• $\sf k-1=0~\Rightarrow~\red{k=1}$

• $\sf k^2+4k-8=0$

$\sf \Delta=4^2-4\cdot1\cdot(-8)$

$\sf \Delta=16+32$

$\sf \Delta=48$

$\sf k=\dfrac{-4\pm\sqrt{48}}{2\cdot1}=\dfrac{-4\pm4\sqrt{3}}{2}$

$\sf k'=\dfrac{-4+4\sqrt{3}}{2}~\Rightarrow~\red{k'=-2\cdot(1-\sqrt{3})}$

$\sf k"=\dfrac{-4-4\sqrt{3}}{2}~\Rightarrow~\red{k"=-2\cdot(1+\sqrt{3})}$

• Para $\sf k=1$:

$\sf log_{2}~x=1$

$\sf x=2^1$

$\sf \green{x=2}$

• Para $\sf k=-2\cdot(1-\sqrt{3})$:

$\sf log_{2}~x=-2\cdot(1-\sqrt{3})$

$\sf \green{x=2^{-2\cdot(1-\sqrt{3})}}$

• Para $\sf k=-2\cdot(1+\sqrt{3})$:

$\sf log_{2}~x=-2\cdot(1+\sqrt{3})$

$\sf \green{x=2^{-2\cdot(1+\sqrt{3})}}$

Logo, $\sf S=\{2^{-2\cdot(1-\sqrt{3})},2^{-2\cdot(1+\sqrt{3})},2\}$