Question

Cho 2 số thực dương \(a\), \(b\) thỏa mãn \(\log_{20}a=\log_8b=\log_{125}\left(5a+12b\right)\). Tính \(P=\dfrac{a+b}{b}\).

Answer 1

Đặt \(log_{20}a=log_8b=log_{125}\left(5a+12b\right)=k\)

\(\Rightarrow\left\{{}\begin{matrix}a=20^k\\b=8^k\\5a+12b=125^k\end{matrix}\right.\)

\(\Rightarrow5.20^k+12.8^k=125^k\)

\(\Rightarrow5.\left(\dfrac{4}{25}\right)^k+12.\left(\dfrac{8}{125}\right)^k=1\)

Đặt \(\left(\dfrac{2}{5}\right)^k=x>0\)

\(\Rightarrow5x^2+12x^3=1\)

\(\Leftrightarrow\left(3x-1\right)\left(4x^2+3x+1\right)=0\)

\(\Leftrightarrow x=\dfrac{1}{3}\)

\(\Rightarrow\left(\dfrac{2}{5}\right)^k=\dfrac{1}{3}\)

\(P=\dfrac{a+b}{b}=\dfrac{a}{b}+1=\dfrac{20^k}{8^k}+1=\left(\dfrac{5}{2}\right)^k+1=3+1=4\)