Bài 2 trang 68 sgk giải tích 12

Bài 2. Tính:

Bài 2. Tính:

a) ${4^{lo{g_2}3}}$;

b) ${27^{lo{g_9}2}}$;

c) ${9^{lo{g_{\sqrt 3 }}2}}$

d) ${4^{lo{g_8}27}}$;.

Hướng dẫn giải:

a) \({4^{lo{g_2}3}} = {\left( {{2^2}} \right)^{lo{g_2}3}} = {\left( {{2^{lo{g_2}3}}} \right)^2} = {3^2} = 9\).

b)

\(\eqalign{
& {27^{lo{g_9}2}} = {\left( {{3^3}} \right)^{lo{g_9}2}} = {3^3}^{lo{g_9}2} = {\left( {{9^{{1 \over 2}}}} \right)^{3lo{g_9}2}} \cr
& = {9^{{1 \over 2}}}^{3lo{g_9}2} = {\left( {{9^{lo{g_9}2}}} \right)^{{3 \over 2}}} = {2^{{3 \over 2}}} = 2\sqrt 2 \cr} \)

c) \({9^{lo{g_{\sqrt 3 }}2}} = {\left( {{{\left( {\sqrt 3 } \right)}^4}} \right)^{lo{g_{\sqrt 3 }}2}} = {\left( {{{\left( {\sqrt 3 } \right)}^{lo{g_{\sqrt 3 }}2}}} \right)^4} = {2^4} = 16\)

d)  Có \({\rm{lo}}{{\rm{g}}_8}{\rm{27 = }}lo{g_{{2^3}}}{3^3} = {3 \over 2}lo{g_2}3 = {\rm{lo}}{{\rm{g}}_2}{\rm{3}}\)

nên \({4^{lo{g_8}27}} = {\left( {{2^2}} \right)^{lo{g_2}3}} = {\left( {{2^{lo{g_2}3}}} \right)^2} = {3^2} = 9\)