log2(3.9) = 9a.
Giải thích
a) \({\log _2}25 = \frac{1}{{{{\log }_{25}}2}} = \frac{1}{b}\).
b) \({\log _2}75 = {\log _2}\left( {25.3} \right) = {\log _2}25 + {\log _2}3 = a + \frac{1}{b}\).
c) log2(3.9) = log233 = 3log23 = 3a.
d) Ta có \({\log _{48600}}25 = \frac{1}{{{{\log }_{25}}48600}} = \frac{1}{{{{\log }_{25}}\left( {{3^5}{{.2}^3}.25} \right)}}\)\( = \frac{1}{{{{\log }_{25}}{3^5} + {{\log }_{25}}{2^3} + {{\log }_{25}}25}}\)
\( = \frac{1}{{5{{\log }_{25}}2.{{\log }_2}3 + 3{{\log }_{25}}2 + 1}}\)\( = \frac{1}{{5ab + 3b + 1}}\).
Suy ra x = 5; y = 3; z = 1. Do đó x + y + z = 9.
Đáp án: a) Đúng; b) Đúng; c) Sai; d) Sai.