Cho log2x = 16. Tính A = log8(log4x2).
Giải thích
C
A = log8(log4x2) = log8(log2x) = log816 = \(\frac{{{{\log }_2}16}}{{{{\log }_2}8}} = \frac{{{{\log }_2}{2^4}}}{{{{\log }_2}{2^3}}} = \frac{4}{3}\).
C
A = log8(log4x2) = log8(log2x) = log816 = \(\frac{{{{\log }_2}16}}{{{{\log }_2}8}} = \frac{{{{\log }_2}{2^4}}}{{{{\log }_2}{2^3}}} = \frac{4}{3}\).