III.- Ejercicios resueltos de logaritmos

	21. Resolver
	$\dpi{100} \bg_black log\, x=\frac{2-log\, x}{log\, x}$

	tenemos:
	$\dpi{100} \bg_black (log\, x)^{2}+log\, x-2=0\: \: \: \: \:;\: \: \: \: log\, x=t$
	$\dpi{100} \bg_black t^{2}+t-2=0\: \: \: \: ;\: \: \: \: t=1\: \: \: \: ;\: \: \: \: t_{1}=-2$
	$\dpi{100} \bg_black log\, x=1\: \: \: \: \: ;\: \: \: \: \: x=10$

	$\dpi{100} \bg_black log\, x=-2\: \: \: \: ;\: \: \: \: x=10^{-2}=\frac{1}{100}$

	22. Resolver
	$\bg_black log\left ( 25-x^{3} \right )-3log\left ( 4-x \right )=0$

	tenemos:
	$\dpi{100} \bg_black log\, (25-x^{3})=log\, (4-x)^{3}$
	$\dpi{100} \bg_black (25-x^{3})=(4-x)^{3}$
	$\dpi{100} \bg_black 25-x^{3}=64-48x+12x^{2}-x^{3}$

	$\dpi{100} \bg_black 12x^{2}-48x+39=0\: \: \: \: ;\: \: \: \: x=2\pm \frac{\sqrt{3}}{2}$

	23. Resolver
	$\bg_black \frac{log\left ( 16-x^{2} \right )}{log\left ( 3x-4 \right )}=2$

	tenemos:
	$\dpi{100} \bg_black log\, (16-x^{2})=2\, log\, (3x-4)$
	$\dpi{100} \bg_black log\, (16-x^{2})=log\, (3x-4)^{2}$
	$\dpi{100} \bg_black 16-x^{2}=(3x-4)^{2}$

	$\dpi{100} \bg_black 10x^{2}-24x=0\: \: \: ;\: \: \: x=0\: \: \: ;\: \: \: x=\frac{12}{5}$

	24. Resolver
	$\dpi{100} \bg_black log\, 2+log\, (11-x^{2})=log\, (5-x)^{2}$

	tenemos:
	$\dpi{100} \bg_black log\,2+log\, (11-x^{2})=2\, log\, (5-x)$
	$\dpi{100} \bg_black log\, \left [ \, 2\, (11-x^{2}) \right ]=log\, (5-x)^{2}$
	$\dpi{100} \bg_black 2(11-x^{2})=(5-x)^{2}$
	$\dpi{100} \bg_black 3x^{2}-10x+3=0$
	$x=3\: \: ;\: \: 11-3^{2}>0\: \: ;\: \: 5-3>0″ alt=”\bg_black x=3\: \: ;\: \: 11-3^{2}>0\: \: ;\: \: 5-3>0″ align=”absmiddle”></td> <td style=$

	$x=\frac{1}{3}\;;\; 11-(\frac{1}{3})^{2}>0\; ;\; 5-\frac{1}{3}>0″ alt=”x=\frac{1}{3}\;;\; 11-(\frac{1}{3})^{2}>0\; ;\; 5-\frac{1}{3}>0″ align=”absmiddle”></td> <td style=$

	25. Resolver
	$\dpi{100} \bg_black (x^{2}-5x+9)log\, 125=3$

	tenemos:
	$\dpi{100} \bg_black log\, 2^{x^{2}-5x+9}+log\, 125=log\, 100$
	$\dpi{100} \bg_black log\, (2^{x^{2}-5x+9}.125)=log\, 1000$
	$\dpi{100} \bg_black 2^{x^{2}-5x+9}.125=1000$
	$\dpi{100} \bg_black 2^{x^{2}-5x+9}=8\: \: \: \: ;\: \: \: \: 2^{x^{2}-5x+9}=2^{3}$
	$\dpi{100} \bg_black x^{2}-5x+9=3\: \: \: ;\: \: \: x^{2}-5x+6=0$
	$\dpi{100} \bg_black x_{1}=2\: \: \: \: \: ;\: \: \: \: \: x_{2}=3$

	26. Resolver
	$\dpi{100} \bg_black (x^{2}-5x+9)log\, 125=3$

$\dpi{100} \bg_black \left\{\begin{matrix} log\, x+log\, y=2\\ \\ x-y=20 \end{matrix}\right.$


	$\dpi{100} \bg_black log\, (xy)=log\, 100\, \, \, ;\, \, \, xy=100\: \: \: ;\: \: \: x=\frac{100}{y}$
	$\dpi{100} \bg_black \frac{100}{y}-y=20\: \: \: ;\: \: \: y^{2}+20y-100=0$
	$\dpi{100} \bg_black y=\frac{-20\pm \sqrt{400+400}}{2}=\frac{-20\pm 20\sqrt{2}}{2}=$ $-10\pm \, 20\sqrt{2}$
	$\dpi{100} \bg_black y=10(\sqrt{2}+1)$ ; $\dpi{100} \bg_black y=10(\sqrt{2}-1)$

	27. Resolver
	$\dpi{100} \bg_black \left\{\begin{matrix} log\, x+log\, y=3\\ \\ log\, x-log\, y=1 \end{matrix}\right.$
	——————————
	$\dpi{100} \bg_black 2\, log\, x$ = 4


	$\dpi{100} \bg_black log\, x=2\: \: \: \: ;\: \: \: \: x=10^{2}\: \: \: \: ;\: \: \: \: x=100$
	$\dpi{100} \bg_black 2+log\, y=3$
	$\dpi{100} \bg_black log\, y=1\: \: \: ;\: \: \: y=10^{1}\: \: \: ;\: \: \: y=10$