Question

20 PUNTOS(CON PROCEDIMIENTO)
es de logaritmos

Answer 1

Respuesta:

a) $-\frac{3}{4}$

Explicación paso a paso:

$\log _{\sqrt{7}}\left(\sqrt[8]{\frac{1}{343}}\right)=y$

$y=\log _{\sqrt{7}}\left(\sqrt[8]{\frac{1}{343}}\right)$

$y=\log _{\sqrt{7}}\left(\left(\frac{1}{343}\right)^{\frac{1}{8}}\right)$

$y=\frac{1}{8}\log _{\sqrt{7}}\left(\frac{1}{343}\right)$

$\log _{\sqrt{7}}\left(\frac{1}{343}\right)\\=\log _{7^{\frac{1}{2}}}\left(\frac{1}{343}\right)\\=\frac{1}{\frac{1}{2}}\log _7\left(\frac{1}{343}\right)\\=2\log _7\left(\frac{1}{343}\right)$

$y=2\cdot \frac{1}{8}\log _7\left(\frac{1}{343}\right)$

$y=\frac{1\cdot \:2}{8}\log _7\left(\frac{1}{343}\right)$

$\frac{1\cdot \:2}{8}\\=\frac{2}{8}\\=\frac{1}{4}$

$y=\frac{1}{4}\log _7\left(\frac{1}{343}\right)$

$\log _7\left(\frac{1}{343}\right)\\=-\log _7\left(343\right)\\=-\log _7\left(7^3\right)\\=-3\log _7\left(7\right)\\=-3\cdot \:1\\=-3$

$y=-3\cdot \frac{1}{4}\\=-\frac{1\cdot \:3}{4}$

$y=-\frac{3}{4}$