Bentuk 1
\(a^{f(x)} = a^{g(x)}\)
\( a^{f(x)} = a^{g(x)} \)
\( \cancel{a}^{f(x)} = \cancel{a}^{g(x)} \)
\( f(x) = g(x) \)
\begin{equation*} \begin{split} & 3^{2x - 1} = 3^{x + 5} \\\\ & \cancel {3}^{2x - 1} = \cancel {3}^{x + 5} \\\\ & 2x - 1 = x + 5 \\\\ & \bbox[5px, border: 2px solid magenta] {x = 6} \end{split} \end{equation*}
Bentuk 2
\(a^{f(x)} = b^{f(x)}\)
\(f(x) = 0\)
\begin{equation*} \begin{split} & 5^{x - 4} = 7^{x - 4} \\\\ & x - 4 = 0 \\\\ & \bbox[5px, border: 2px solid magenta] {x = 4} \end{split} \end{equation*}
Bentuk 3
\(a^{f(x)} = b \: ^{g(x)}\)
Menambahkan \(\log\) pada kedua ruas.
\( \log a^{f(x)} = \log b \: ^{g(x)}\)
\( f(x) \:.\: \log a = g(x) \:.\: \log b \)
\begin{equation*} \begin{split} & 5^{x - 3} = 2^{x + 1} \\\\ & \log 5^{x - 3} = \log 2^{x + 1} \\\\ & (x - 3) \:.\: \log 5 = (x + 1) \:.\: \log 2 \\\\ & x \:.\: \log 5 - 3 \log 5 = x \:.\: \log 2 + \log 2 \\\\ & x \:.\: \log 5 - x \:.\: \log 2 = 3 \log 5 + \log 2 \\\\ & x \:.\: (\log 5 - \log 2) = 3 \log 5 + \log 2 \\\\ & \bbox[5px, border: 2px solid magenta] {x = \frac {3 \log 5 + \log 2}{\log 5 - \log 2}} \end{split} \end{equation*}
Bentuk 4
Bentuk penfaktoran
\begin{equation*} \begin{split} & 2^{2x} - 6 \:.\: 2^{x + 1} + 32 = 0 \\\\ & (2^x)^2 - 6 \:.\: 2^x \cdot 2^1 + 32 = 0 \\\\ & (2^x)^2 - 12 \:.\: 2^x + 32 = 0 \\\\ & {\color {blue} \text{misalkan } m = 2^x} \\\\ & m^2 - 12 m + 32 = 0 \\\\ & (m - 4)(m - 8) = 0 \end{split} \end{equation*}
Faktor 1
\begin{equation*} \begin{split} & m - 4 = 0 \\\\ & m = 4 \\\\ & 2^x = 2^2 \\\\ & \bbox[5px, border: 2px solid magenta] {x = 2} \end{split} \end{equation*}
Faktor 2
\begin{equation*} \begin{split} & m - 8 = 0 \\\\ & m = 8 \\\\ & 2^x = 2^3 \\\\ & \bbox[5px, border: 2px solid magenta] {x = 3} \end{split} \end{equation*}
Bentuk 5
\(f(x)^{h(x)} = g(x)^{h(x)}\)
Solusi 1
\begin{equation*} \begin{split} f(x)^{h(x)} & = g(x)^{h(x)}\\\\ f(x)^{\cancel{h(x)}} & = g(x)^{\cancel{h(x)}}\\\\ f(x) & = g(x) \end{split} \end{equation*}
Solusi 2
\(h(x) = 0 \)
Dengan syarat nilai \(f(x) \neq 0\) dan \(g(x) \neq 0\)
\((2x + 3)^{x - 1} = (3x + 5)^{x - 1}\)
Solusi 1
\begin{equation*} \begin{split} & (2x + 3)^{x - 1} = (3x + 5)^{x - 1} \\\\ & (2x + 3)^{\cancel {x - 1}} = (3x + 5)^{\cancel {x - 1}} \\\\ & 2x + 3 = 3x + 5 \\\\ & \bbox[5px, border: 2px solid magenta] {x = -2} \end{split} \end{equation*}
Solusi 2
\begin{equation*} \begin{split} & x - 1 = 0 \\\\ & \bbox[5px, border: 2px solid magenta] {x = 1} \end{split} \end{equation*}
Uji \(x = 1\) apakah memenuhi \(2x + 3 \neq 0\) dan \(3x + 5 \neq 0\)
\begin{equation*} \begin{split} & 2 \:.\: 1 + 3 \neq 0 \\\\ & 3 \:.\: 1 + 5 \neq 0 \end{split} \end{equation*}
Bentuk 6
\(h(x)^{f(x)} = h(x)^{g(x)}\)
Solusi 1
\begin{equation*} \begin{split} h(x)^{f(x)} & = h(x)^{g(x)} \\\\ \cancel {h(x)}^{f(x)} & = \cancel {h(x)}^{g(x)} \\\\ f(x) & = g(x) \end{split} \end{equation*}
Solusi 2
\(h(x) = 1 \)
Solusi 3
\(h(x) = 0 \)
Dengan syarat nilai \(f(x) > 0\) dan \(g(x) > 0\)
Solusi 4
\(h(x) = -1 \)
Dengan syarat nilai \(f(x)\) dan \(g(x)\) keduanya bilangan ganjil atau keduanya bilangan genap
\((x - 5)^{x^2 - 4} = (x - 5)^{2 - x}\)
Solusi 1
\begin{equation*} \begin{split} & (x - 5)^{x^2 - 4} = (x - 5)^{2 - x} \\\\ & \cancel{(x - 5)}^{x^2 - 4} = \cancel{(x - 5)}^{2 - x} \\\\ & x^2 - 4 = 2 - x \\\\ & x^2 + x - 6 = 0 \\\\ & (x + 3)(x - 2) = 0 \\\\ & \bbox[5px, border: 2px solid magenta] {x = -3 \text{ atau } x = 2} \end{split} \end{equation*}
Solusi 2
\begin{equation*} \begin{split} & (x - 5)^{x^2 - 4} = (x - 5)^{2 - x} \\\\ & x - 5 = 1 \\\\ & \bbox[5px, border: 2px solid magenta] {x = 6} \end{split} \end{equation*}
Solusi 3
\begin{equation*} \begin{split} & (x - 5)^{x^2 - 4} = (x - 5)^{2 - x} \\\\ & x - 5 = 0 \\\\ & \bbox[5px, border: 2px solid magenta] { x = 5} \end{split} \end{equation*}
Uji \(x = 5\) apakah memenuhi \(x^2 - 4 > 0\) dan \(2 - x > 0\).
\begin{equation*} \begin{split} & x^2 - 4 = (5)^2 - 4 = 21 \quad {\color {red} > 0}\\\\ & 2- x = 2 - 5 = -3 \quad {\color {red} < 0} \end{split} \end{equation*}
Karena \(2 - x < 0\), maka \(x = 5\) tidak memenuhi syarat.
Solusi 4
\begin{equation*} \begin{split} & (x - 5)^{x^2 - 4} = (x - 5)^{2 - x} \\\\ & x - 5 = -1 \\\\ & \bbox[5px, border: 2px solid magenta] {x = 4} \end{split} \end{equation*}
Uji \(x = 5\) apakah memenuhi \(x^2 - 4\) dan \(2 - x\) keduanya bilangan ganjil atau bilangan genap.
\begin{equation*} \begin{split} & x^2 - 4 = (4)^2 - 4 = 12 \quad {\color {red}\text{genap}}\\\\ & 2 - x = 2 - 4 = -2 \quad {\color {red}\text{genap}} \end{split} \end{equation*}
Karena \(x^2 - 4\) dan \(2 - x\) bernilai genap, maka \(x = 4\) memenuhi syarat.
HP = \(\{ -3,2,4,6 \}\)
SOAL LATIHAN