Tính: 4log2(3)

Giải Toán 12 Bài 3: Lôgarit

Bài 2 trang 68 Toán lớp 12 Giải tích:

a) $4^{{\log }_{2}3}$;

b) ${27}^{{\log }_{9}2}$;

c) $9^{{\log }_{\sqrt{3}}2}$;

d) $4^{{\log }_{8}27}$.

Lời giải:

a) $4^{{\log }_{2}3}={\left(2^2\right)}^{{\log }_{2}3}={\left(2^{{\log }_{2}3}\right)}^2=3^2=9$

$\mathbf{b}\mathbf{)}{27}^{{\log }_{9}2}={\left(3^3\right)}^{{\log }_{9}2}=3^{3.{\log }_{3^2}2}=3^{\frac{3}{2}.{\log }_{3}2}={\left(3^{{\log }_{3}2}\right)}^{\frac{3}{2}}=2^{\frac{3}{2}}={\left(\sqrt{2}\right)}^3=2\sqrt{2}$

$\mathbf{c}\mathbf{)}{9}^{{\log }_{\sqrt{3}}2}={\left({\left(\sqrt{3}\right)}^4\right)}^{{\log }_{\sqrt{3}}2}={\left({\sqrt{3}}^{{\log }_{\sqrt{3}}2}\right)}^4=2^4=16$

d) $4^{{\log }_{8}27}={\left(2^2\right)}^{{\log }_{2^3}{3}^3}={\left(2^2\right)}^{\frac{1}{3}.3{\log }_{2}3}$

$={\left(2^2\right)}^{{\log }_{2}3}={\left(2^{{\log }_{2}3}\right)}^2=3^2=9$