Question

Biết rằng \({10^\alpha } = 2;{10^\beta } = 5\).

Tính \({10^{\alpha  + \beta }};{10^{\alpha  - \beta }};{10^{2\alpha }};{10^{ - 2\alpha }};{1000^\beta };0,{01^{2\alpha }}\).

Answer 1

Ta có: 

\(10^{\alpha}=2\Rightarrow\alpha=log_{10}2\)

\(10^{\beta}=5\Rightarrow\beta=log_{10}5\)

Kết quả:

\(10^{\alpha+\beta}=10^{log_{10}2+log_{10}5}=10\)

\(10^{2\cdot log_{10}2}=4\)

\(1000^{log_{10}5}=125\)

\(0,01^{2\cdot log_{10}2}=\dfrac{1}{16}\)