Giải Bài tập 6.35 trang 26 sgk Toán 11 tập 2 Kết nối

$B= \log_{a}\left(\frac{a^{2}\sqrt[3]{a}\sqrt[5]{a^{4}}\sqrt[5]{a^{4}}}{\sqrt[4]{5}}\right) + a^{2\log_{a}\frac{\sqrt{105}}{30}} $

$= \log_{a}(a^{2}) + \log_{a}(\sqrt[3]{a}) + \log_{a}(\sqrt[5]{a^{4}}) + \log_{a}(\sqrt[5]{a^{4}}) - \log_{a}(\sqrt[4]{5}) + a^{2\log_{a}\frac{\sqrt{105}}{30}}$

$= 2 + \frac{1}{3}\log_{a}a + \frac{4}{5}\log_{a}a - \frac{1}{4}\log_{a}5 + a^{2\log_{a}\frac{\sqrt{105}}{30}}$

$ = 2 + \frac{1}{3} + \frac{4}{5} - \frac{1}{4}\log_{a}5 + a^{2\log_{a}\frac{\sqrt{105}}{30}} $

$=\frac{31}{15} - \frac{1}{4}\log_{a}5 + a^{2\log_{a}\frac{\sqrt{105}}{30}}$

Tính giá trị của $a^{2\log_{a}\frac{\sqrt{105}}{30}}$:

$a^{2\log_{a}\frac{\sqrt{105}}{30}}=a^{\log_{a}\left(\left(\frac{\sqrt{105}}{30}\right)^{2}\right)}$

$= \left(\frac{\sqrt{105}}{30}\right)^{2}= \frac{105}{900} = \frac{7}{60}$

Vậy ta có:

$B = \frac{31}{15} - \frac{1}{4}\log_{a}5 + a^{2\log_{a}\frac{\sqrt{105}}{30}}$ 

$= \frac{31}{15} - \frac{1}{4}\log_{a}5 + \frac{7}{60} = \frac{205 - 3\log_{a}5}{60}$