# Persamaan dan Pertidaksamaan Eksponen

### Persamaan eksponen

##### $$a^{f(x)} = a^{g(x)}$$

$$a^{f(x)} = a^{g(x)}$$

$$\cancel{a}^{f(x)} = \cancel{a}^{g(x)}$$

$$f(x) = g(x)$$

\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*}

##### $$a^{f(x)} = b^{f(x)}$$

$$f(x) = 0$$

\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*}

##### $$a^{f(x)} = b \: ^{g(x)}$$

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 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*}

##### $$f(x)^{h(x)} = g(x)^{h(x)}$$

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*}

##### $$h(x)^{f(x)} = h(x)^{g(x)}$$

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

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