Giả sử p và q là hai số dương sao cho {log _{16}}p = {log _{20}}q = {log _{25}}(p+q). Tìm giá trị p/q

Đặt $t = {\log _{16}}p = {\log _{20}}q = {\log _{25}}\left( {p + q} \right) \Rightarrow \left\{ \begin{array}{l}p = {16^t}\\q = {20^t}\\p + q = {25^t}\end{array} \right. \Rightarrow \frac{p}{q} = {\left( {\frac{4}{5}} \right)^t}$

Ta có:

$\begin{array}{l}p + q = {25^t} \Leftrightarrow {16^t} + {20^t} = {25^t} \Leftrightarrow {\left( {\frac{4}{5}} \right)^t} + 1 = {\left( {\frac{5}{4}} \right)^t}\\ \Leftrightarrow {\left( {\frac{4}{5}} \right)^{2t}} + {\left( {\frac{4}{5}} \right)^t} - 1 = 0 \Leftrightarrow \left[ \begin{array}{l}{\left( {\frac{4}{5}} \right)^t} = \frac{{ - 1 + \sqrt 5 }}{2}\\{\left( {\frac{4}{5}} \right)^t} = \frac{{ - 1 - \sqrt 5 }}{2}\end{array} \right.\end{array}$

$\Rightarrow {\left( {\frac{4}{5}} \right)^t} = \frac{{ - 1 + \sqrt 5 }}{2} \Leftrightarrow \frac{p}{q} = \frac{1}{2}\left( { - 1 + \sqrt 5 } \right).$