\(7^{2x} + 7^{1+x}\:-\: 98 = 0\)
\((7^{x})^2 + 7\cdot 7^x \:-\: 98 = 0\)
\(\text{Misal: } 7^x = p\)
\(p^2 + 7p \:-\:98 = 0\)
\((p + 14)(p \:-\: 7) = 0\)
\(p_1 = – 14\)
\(7^x = -14\:\:\:\:\:\color{red}\text{tidak ada solusi}\)
\(\text{nilai } 7^x \text{ selalu positif untuk setiap } x\)
\(p_2 = 7\)
\(7^x = 7\)
\(x = 1\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace 1 \rbrace}\)
\(2^{x+1} + 3(\sqrt{2})^x \:-\: 14 = 0\)
\(2\cdot 2^x + 3\cdot 2^{\frac{1}{2}x}\:-\: 14 = 0\)
\(\text{Misal: } 2^{\frac{1}{2}x} = p\)
\(2p^2 + 3p \:-\: 14 = 0\)
\((2p + 7)(p\:-\:2) = 0\)
\(2p + 7 = 0\)
\(2p = -7\)
\(p = -\dfrac{7}{2}\)
\(2^{\frac{1}{2}x} = -\dfrac{7}{2}\:\:\:\:\:\color{red}\text{tidak memenuhi}\)
Nilai \(2^{\frac{1}{2}x} \) selalu positif untuk setiap \(x\)
\(p\:-\:2 =0\)
\(p = 2\)
\(2^{\frac{1}{2}x} = 2^1\)
\(\frac{1}{2}x = 1\)
\(x = 2\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace 2 \rbrace}\)
\(3^x + 3^{3-x} – 12 = 0\)
\(3^x + \dfrac{3^3}{3^x} – 12 = 0\)
\(\text{Misal: } 3^x = p\)
\(p + \dfrac{27}{p} – 12 = 0\)
\(\color{blue}\text{kalikan kedua ruas dengan } p\)
\(p^2 + 27 – 12p = 0\)
\(p^2 \:-\: 12 p + 27 = 0\)
\((p \:-\: 9)(p \:-\: 3) = 0\)
\(p_1 = 9\)
\(3^x = 9\)
\(3^x = 3^2\)
\(x = 2\)
\(p_2 = 3\)
\(3^x = 3\)
\(3^x = 3^1\)
\(x = 1\)
\(\bbox[yellow, 5px]{\text{HP} = \lbrace 1, \:2 \rbrace}\)
