\(27^{11}=\left(3^3\right)^{11}=3^{33};81^8=\left(3^4\right)^8=3^{32}\)
mà 33>32
nên \(27^{11}>81^8\)
\(625^5=\left(5^4\right)^5=5^{20};125^7=\left(5^3\right)^7=5^{21}\)
mà 20<21
nên \(625^5< 125^7\)
\(5^{36}=\left(5^3\right)^{12}=125^{12};11^{24}=\left(11^2\right)^{12}=121^{12}\)
mà 125>121
nên \(5^{36}>11^{24}\)
\[
27 = 3^3 \implies 27^{11} = (3^3)^{11} = 3^{33}
\]
\[
81 = 3^4 \implies 81^{8} = (3^4)^{8} = 3^{32}
\]
- \(27^{11} = 3^{33}\)
- \(81^{8} = 3^{32}\)
\[
27^{11} > 81^{8}
\]
\[
625 = 5^4 \implies 625^{5} = (5^4)^{5} = 5^{20}
\]
\[
125 = 5^3 \implies 125^{7} = (5^3)^{7} = 5^{21}
\]
- \(625^{5} = 5^{20}\)
- \(125^{7} = 5^{21}\)
Vì \(5^{21} > 5^{20}\), nên:
\[
125^{7} > 625^{5}
\]
\[
\log_{10} (5^{36}) = 36 \log_{10} (5) \approx 36 \times 0.6990 = 25.164
\]
\[
\log_{10} (11^{24}) = 24 \log_{10} (11) \approx 24 \times 1.0414 = 24.98
\]
\[
25.164 > 24.98
\]
Vì vậy:
\[
5^{36} > 11^{24}
\]
