Bài 2 trang 68 sgk giải tích 12: Bài 3. Lôgarit...

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}});.

Giải:

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

b)

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

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

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

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