Chứng minh 2^100 < 10^31

Giải thích các bước giải:

Ta có:

`@``2^100=2^31 . 2^69=2^31 . 2^63 . 2^6=2^31 . (2^9)^7 . 2^6=2^31 . 512^7. 64`

`@``10^31=2^31 . 5^31 =2^31 . 5^28 . 5^3=2^31 . (5^4)^7 . 5^3=2^31 . 625^7 . 125`

Vì 

`@``2^31=2^31`

`@``512^7<625^7`

`@``64<125`

Nên `2^31 . 512^7. 64<=2^31 . 625^7 . 125`

Hay `2^100<10^31`

Vậy `2^100<10^31`

Đáp án:

$2^{100}=2^{31}.2^{69}\\10^{31}=\left(2.5\right)^{31}=2^{31}.5^{31}\\2^{69}=2^{68}.2=\left(2^2\right)^{34}.2=4^{34}.2\\=4^{31}.4^3.2=4^{31}.128>4^{31}.125=4^{31}.5^3\\5^{31}=\dfrac{5^{31}}{4^{31}}\!\cdot\!4^{31}>\dfrac{5^{31}}{5^{29}}\!\cdot\!4^{31}(4^{31}<5^{29})\\4^{31}<4^{32}<5^{32}\\5^{29}<5^{32}\Rightarrow 4^{31}<4^{32}<5^{29}\\\Rightarrow 4^{31}<5^{29}(\rm{đpcm})\\\Rightarrow 5^{31}>\dfrac{5^{31}}{5^{29}}\!\cdot\!4^{31}=5^3.4^{31}=125.4^{31}>2^{69}\\\Rightarrow 5^{31}>2^{69}\\\Rightarrow 2^{31}.2^{69}<2^{31}.5^{31}\Rightarrow 2^{100}<10^{31}$

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