Tính:
a) \({4^{log_{2}3}}\); b) \({27^{log_{9}2}}\);
c) \({9^{log_{{\sqrt 3 }}2}}\) d) \({4^{log_{8}27}}\);.
+) Công thức lũy thừa: \({\left( {{a^m}} \right)^n} = {a^{m.n}};\;\;\sqrt {{a^m}} = {a^{\frac{m}{2}}}.\)
+) Sử dụng công thức logarit: \({a^{{{\log }_a}b}} = b; \, \, {\log _a}{b^n} = n{\log _a}b;\;\;{\log _{{a^m}}}b = \frac{1}{m}{\log _a}b .\)
Lời giải chi tiết
a) \({4^{lo{g_2}3}} = {\left( {{2^2}} \right)^{lo{g_2}3}} = {\left( {{2^{lo{g_2}3}}} \right)^2} = {3^2} = 9\).
\(\eqalign{ b) & {27^{lo{g_9}2}} = {\left( {{3^3}} \right)^{lo{g_9}2}} = {\left( {{9^{{1 \over 2}}}} \right)^{3lo{g_9}2}} \cr & = {\left( {{9^{lo{g_9}2}}} \right)^{{3 \over 2}}} = {2^{{3 \over 2}}} = 2\sqrt 2 \cr} \)
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 \({\rm{lo}}{{\rm{g}}_8}{\rm{27 = }}lo{g_{{2^3}}}{3^3} = {3 \over 3}lo{g_2}3 = {\rm{lo}}{{\rm{g}}_2}{\rm{3}}\)
nên \({4^{lo{g_8}27}} = {\left( {{2^2}} \right)^{lo{g_2}3}} = {\left( {{2^{lo{g_2}3}}} \right)^2} = {3^2} = 9\).
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