\((2x^2 - 13x + 15)^{x - 3} = (x^2 - 4x + 1)^{x - 3}\)
Solusi
Solusi 1
\begin{equation*} \begin{split} (2x^2 - 13x + 15)^{x - 3} & = (x^2 - 4x + 1)^{x - 3} \\\\ 2x^2 - 13x + 15)^{\cancel{x - 3}} & = (x^2 - 4x + 1)^{\cancel{x - 3}} \\\\ 2x^2 - 13x + 15 & = x^2 - 4x + 1 \end{split} \end{equation*}
Solusi 2
\(x - 3 = 0\)
Dengan syarat nilai \(2x^2 - 13x + 15 \neq 0\) dan \(x^2 - 4x + 1 \neq 0\)
Solusi 1
\begin{equation*}
\begin{split}
2x^2 - 13x + 15 & = x^2 - 4x + 1 \\\\
x^2 + 9x + 14 & = 0 \\\\
(x + 7)(x + 2) & = 0 \\\\
x = -7 \text{ atau } x & = -2
\end{split}
\end{equation*}
Solusi 2
\begin{equation*}
\begin{split}
x - 3 & = 0 \\\\
x & = 3
\end{split}
\end{equation*}
Nilai \(x = 3\) harus diuji dengan syarat \(2x^2 - 13x + 15 \neq 0\) dan \(x^2 - 4x + 1 \neq 0\).
Uji nilai \(x = 3\)
\begin{equation*} \begin{split} 2x^2 - 13x + 15 & = 2(3)^2 - 13(3) + 15 \\\\ 2x^2 - 13x + 15 & = -6 \quad {\color {red} \neq 0} \end{split} \end{equation*}
\begin{equation*} \begin{split} x^2 - 4x + 1 & = (3)^2 - 4(3) + 1 \\\\ x^2 - 4x + 1 & = -2 \quad {\color {red} \neq 0} \end{split} \end{equation*}
Karena nilai \(2x^2 - 13x + 15 \neq o\) dan \(x^2 - 4x + 1 \neq o\), maka \(x = 3\) memenuhi syarat sebagai solusi.
HP = \(\{ -7,-2,3 \}\)