\(3^{x^2 – 3} = 3^1\)
\(x^2 \:-\: 3 = 1\)
\(x^2 \:-\: 4 = 0\)
\(x^2\:-\: 2^2 = 0, \:\:\:\:\:\color{blue} \text{ingat } a^2 \:-\: b^2 = (a + b)(a\:-\: b)\)
\((x + 2)(x \:-\: 2) = 0\)
\(x + 2 = 0 \rightarrow x = -2\)
\(x – 2 = 0 \rightarrow x = 2\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace -2, 2 \rbrace}\)
\(2^{x^2 + 5x + 9} = 2^3\)
\(x^2 + 5x + 9 = 3\)
\(x^2 + 5x + 6 = 0\)
\((x + 2)(x + 3) = 0\)
\(x + 2 = 0 \rightarrow x = -2\)
\(x + 3 = 0 \rightarrow x = -3\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace -2,3 \rbrace}\)
\((5^2)^{x^2 + 7x \:-\: 6} = 5^4\)
\(5^{2x^2 + 14x \:-\: 12} = 5^4\)
\(2x^2 + 14x \:-\: 12 = 4\)
\(2x^2 + 14x\:-\: 16= 0\)
\(\color{blue} \text{ bagi kedua ruas dengan 2}\)
\(x^2 + 7x \:-\: 8= 0\)
\((x + 8)(x \:-\: 1) = 0\)
\(x + 8 = 0 \rightarrow x = -8\)
\(x – 1 = 0 \rightarrow x = 1\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace -8, 1 \rbrace}\)
Tentukan himpunan penyelesaian persamaan \(\left(\dfrac{1}{2}\right)^{-2x^2\:-\:3x} = 2^{14}\)
\((2^{-1})^{-2x^2\:-\:3x} = 2^{14}\)
\(2^{2x^2 + 3x} = 2^{14}\)
Karena bilangan pokok pada ruas kiri sama dengan ruas kanan maka tinggal samakan pangkatnya.
\(2x^2 + 3x = 14\)
\(2x^2 + 3x \:-\:14 = 0\)
Selanjutnya faktorkan
\((x\:-\:2)(2x + 7) = 0\)
\(x\:-\:2 = 0 \rightarrow x = 2\)
\(2x + 7 = 0 \rightarrow x = -\dfrac{7}{2}\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace -\dfrac{7}{2}, 2\rbrace}\)
