Giải thích các bước giải:
$d.64^{2log_27}=(2^6)^{2log_27}=2^{12log_27}=(2^{log_27})^{12}=7^{12}$
$f.(0.25)^{3log_25}=(\dfrac{1}{2^2})^{3log_25}=\dfrac{1}{2^{3log_25}}=\dfrac{1}{(2^{log_25})^3}=\dfrac{1}{5^3}=\dfrac{1}{125}$
$e.4^{2+log_23}=4^2.4^{log_23}=16.(2^{2})^{log_23}=16.(2^{log_23})^2=16.3^2=16.9=144$