Bài 18 trang 58

Đề bài

Cho $a>0;a\ne 1;a^{\frac{3}{5}}=b$.

a) Viết $a^6;a^{3}b;\frac{a^9}{b^9}$ theo lũy thừa cơ số b.

b) Tính ${\log }_{a}b;{\log }_a\left(a^2{b}^5\right);{\log }_{\sqrt[5]{5a}}\left(\frac{a}{b}\right)$.

Lời giải chi tiết

a) $a^6=a^{\frac{30}{5}}={\left(a^{\frac{3}{5}}\right)}^{10}=b^{10}$

$a^{3}b=a^{\frac{15}{5}}b={\left(a^{\frac{3}{5}}\right)}^{5}b=b^5.b=b^6$

$\left(\frac{a^9}{b^9}\right)={\left(\frac{a}{b}\right)}^9={\left(\frac{a}{a^{\frac{3}{5}}}\right)}^9={\left(a^{\frac{2}{5}}\right)}^9=a^{\frac{18}{5}}={\left(a^{\frac{3}{5}}\right)}^6=b^6$

b) ${\log }_{a}b={\log }_a{a}^{\frac{3}{5}}=\frac{3}{5}$

${\log }_a\left(a^2{b}^5\right)={\log }_a\left(a^2.{\left(a^{\frac{3}{5}}\right)}^5\right)={\log }_a\left(a^2.a^3\right)={\log }_a\left(a^5\right)=5$

${\log }_{\sqrt[5]{5a}}\left(\frac{a}{b}\right)={\log }_{a^{\frac{1}{5}}}\left(\frac{a}{a^{\frac{3}{5}}}\right)=5{\log }_a{a}^{\frac{2}{5}}=2$