Đặt a = log 27 ^ 5

10/86

Đặt \(a = {\rm{lo}}{{\rm{g}}_{27}}5,b = {\rm{lo}}{{\rm{g}}_8}7,c = {\rm{lo}}{{\rm{g}}_2}3\). Khi đó \({\rm{lo}}{{\rm{g}}_{12}}35\) bằng:

\(\frac{{3ac + 3b}}{{c + 1}}\).

\(\frac{{2ac + 3b}}{{c + 3}}\).

\(\frac{{3ac + 3b}}{{c + 2}}\).

\(\frac{{2ac + 3b}}{{c + 2}}\)

Giải thích

Đáp án đúng là C

Phương pháp giải

Sử dụng công thức, tính chất logarit.

Lời giải

\({\rm{lo}}{{\rm{g}}_{12}}35 = {\rm{lo}}{{\rm{g}}_{12}}5 + {\rm{lo}}{{\rm{g}}_{12}}7 = \frac{1}{{{\rm{lo}}{{\rm{g}}_5}12}} + \frac{1}{{{\rm{lo}}{{\rm{g}}_7}12}} = \frac{1}{{{\rm{lo}}{{\rm{g}}_5}\left( {{{3.2}^2}} \right)}} + \frac{1}{{{\rm{lo}}{{\rm{g}}_7}\left( {{{3.2}^2}} \right)}}\)

\( = \frac{1}{{{\rm{lo}}{{\rm{g}}_5}3 + 2{\rm{lo}}{{\rm{g}}_5}2}} + \frac{1}{{{\rm{lo}}{{\rm{g}}_7}3 + 2{\rm{lo}}{{\rm{g}}_7}2}}\).

Ta có:

\(a = {\rm{lo}}{{\rm{g}}_{27}}5 = {\rm{lo}}{{\rm{g}}_{{3^3}}}5 = \frac{1}{3}{\rm{lo}}{{\rm{g}}_3}5 \Rightarrow {\rm{lo}}{{\rm{g}}_5}3 = \frac{1}{{3a}}\).

\(b = {\rm{lo}}{{\rm{g}}_8}7 = {\rm{lo}}{{\rm{g}}_{{2^3}}}7 = \frac{1}{3}{\rm{lo}}{{\rm{g}}_2}7 \Rightarrow {\rm{lo}}{{\rm{g}}_7}2 = \frac{1}{{3b}}\).

\(c = {\rm{lo}}{{\rm{g}}_2}3 = \frac{{{\rm{lo}}{{\rm{g}}_5}3}}{{{\rm{lo}}{{\rm{g}}_5}2}} = \frac{{\frac{1}{{3a}}}}{{{\rm{lo}}{{\rm{g}}_5}2}} \Rightarrow {\rm{lo}}{{\rm{g}}_5}2 = \frac{1}{{3ac}}\).

\({\rm{lo}}{{\rm{g}}_7}3 = {\rm{lo}}{{\rm{g}}_7}2.{\rm{lo}}{{\rm{g}}_2}3 = \frac{c}{{3b}}\).

Vậy \({\rm{lo}}{{\rm{g}}_{12}}35 = \frac{1}{{\frac{1}{{3a}} + \frac{2}{{3ac}}}} + \frac{1}{{\frac{c}{{3b}} + \frac{2}{{3b}}}} = \frac{{3ac}}{{c + 2}} + \frac{{3b}}{{c + 2}} = \frac{{3ac + 3b}}{{c + 2}}\).