Menyelesaikan Persamaan Eksponen

Kali ini kita akan menyelesaikan persamaan eksponen. Persamaan eksponen memiliki berbagai variasi soal dan cara untuk mengerjakan.

Persamaan Eksponen dengan Basis Sama

Jika \(a\) adalah bilangan bulat, \(a \neq 0\) maka berlaku hubungan \(a^{f(x)}=a^p\) sehingga \(f(x)=p\).

Contoh:
\[a.\quad 2^x = 32\] \[\quad \quad 2^x = 2^5\] \[\quad \quad x = 5\]

\[b.\quad 2^{x+1}.3^{x-1} = 24\] \[\quad \quad 2^x.2.3^x.\frac{1}{3}=24\] \[\quad \quad (2.3)^x = \frac{24\times 3}{2}\] \[\quad \quad 6^x = 36\] \[\quad \quad 6^x = 6^2 \] \[\quad \quad x = 2\]

\[c.\quad \left(\frac{1}{3}\right)^3 \sqrt{3^{2x+1}} = 27\] \[\quad \quad 3^{-3}.3^{\frac{1}{2}(2x+1)} = 3^3\] \[\quad \quad 3^{\frac{1}{2}(2x+1)} = 3^6\] \[\quad \quad \frac{1}{2}(2x+1)=6 \] \[\quad \quad 2x+1=12 \] \[\quad \quad x=\frac{11}{2}\]

\[d.\quad \sqrt[4]{(0,2)^x} = 25^{x+1}\] \[\quad \quad \left(\frac{1}{5}\right)^{\frac{x}{4}} = 5^{2x+2}\] \[\quad \quad 5^{\frac{-x}{4}} = 5^{2x+2}\] \[\quad \quad \frac{-x}{4} = 2x+2\] \[\quad \quad -x = 8x+8 \] \[\quad \quad x=\frac{-8}{9}\]

Persamaan Eksponen dengan Basis Berbeda

Untuk menyelesaikan persamaan eksponen ini dibutuhkan \(\log\) atau \(ln\)

Contoh:
\[a.\quad 2^{x+3} = 5^{2x}\] \[\quad \quad \ln2^{x+3} = \ln5^{2x}\] \[\quad \quad (x+3)\ln2 = (2x)\ln5\] \[\quad \quad x\ln2 + 3\ln2 = 2x\ln5\] \[\quad \quad 3\ln2 = 2x\ln5 – x\ln2\] \[\quad \quad 3\ln2 = x(2\ln5 – \ln2)\] \[\quad \quad x = \frac{3\ln2}{2\ln5-\ln2}\] \[\quad \quad x = \frac{\ln8}{\ln25-\ln2}\]

\[b.\quad 2^{x+3} = 10^{x+3}\] \[\quad \quad \log2^{x+3} = \log10^{x+3}\] \[\quad \quad (x+3)\log2 = (x+3)\log10\] \[\quad \quad x\log2 + 3\log2 = x + 3\] \[\quad \quad 3\log2 – 3 = x – x\log2\] \[\quad \quad 3(\log2-1) = x(1-\log2)\] \[\quad \quad x = \frac{3(\log2 – 1)}{1-log2}\] \[\quad \quad x = \frac{3(\log2-1)}{-(\log2-1)} = -3\]

Persamaan Kuadrat Eksponen

Ini adalah bentuk persamaan kuadrat, hanya saja variabelnya adalah eksponen, untuk menyelesaikan persamaan eksponen ini dibutuhkan pemisalan variabel, seperti \(y = 3^x\) atau \(y = 2^x\), dsb..

Contoh:
\[a.\quad 2^{2x-1} – 9.2^{x-2} + 1 = 0\] \[\quad \quad \frac{1}{2}.2^{2x} – 9.\frac{1}{4}.2^x + 1 = 0\] \[\quad \quad \frac{1}{2}.\left(2^x\right)^2 – \frac{9}{4}.(2^x) + 1 = 0\quad y = 2^x\] \[\quad \quad \frac{1}{2}y^2 – \frac{9}{4}y + 1 = 0\] \[\quad \quad \quad (y – \frac{1}{2})(\frac{1}{2}y-2)\] \[\quad \quad \quad y = \frac{1}{2} \lor y = 4\] \[\quad \quad \quad y = \frac{1}{2} \quad \Rightarrow \quad 2^x=\frac{1}{2} \quad \quad x = -1\] \[\quad \quad \quad y = 4 \quad \Rightarrow \quad 2^x=4 \quad \quad x = 2\]

\[b.\quad 3^{2x+1} + 9 = 3^{x+3} + 3^x\] \[\quad \quad 3.\left(3^x\right)^2 + 9 = 27.(3^x) + 3^x\] \[\quad \quad 3.\left(3^x\right)^2 + 9 = 28.(3^x)\quad y=3^x\] \[\quad \quad 3y^2 – 28y + 9 = 0\] \[\quad \quad \quad (y – 9)(3y – 1)\] \[\quad \quad \quad y = 9 \lor y=\frac{1}{3}\] \[\quad \quad \quad y = \frac{1}{3} \quad \Rightarrow \quad 3^x = \frac{1}{3} \quad \quad x=-1\] \[\quad \quad \quad y = 9 \quad \Rightarrow \quad 3^x = 9 \quad \quad x=2\]

\[c.\quad 9^{x+1} + 81 = 246.3^x\] \[\quad \quad 9.\left(3^x\right)^2 + 81 = 246.(3^x) \quad y=3^x\] \[\quad \quad 9y^2 – 246y + 81 = 0 \quad (\div 3)\] \[\quad \quad 3y^2 – 82y + 27 =0\] \[\quad \quad \quad (y – 27)(3y – 1)\] \[\quad \quad \quad y = 27 \lor y = \frac{1}{3}\] \[\quad \quad \quad y = 27 \quad \Rightarrow \quad 3^x = 27 \quad \quad x = 3\] \[\quad \quad \quad y = \frac{1}{3} \quad \Rightarrow \quad 3^x = \frac{1}{3} \quad \quad x=-1\]

\[d. \quad 5.2^x = 2.4^x + 2\] \[\quad \quad 5.2^x = 2.\left(2^x\right)^2 + 2 \quad y = 2^x\] \[\quad \quad 5y = 2y^2 + 2\] \[\quad \quad 2y^2 – 5y + 2 = 0\] \[\quad \quad \quad (y – 2)(2y – 1)\] \[\quad \quad \quad y = 2 \lor y = \frac{1}{2}\] \[\quad \quad \quad y = 2 \quad \Rightarrow \quad 2^x = 2 \quad \quad x = 1\] \[\quad \quad \quad y = \frac{1}{2} \quad \Rightarrow \quad 2^x = \frac{1}{2} \quad \quad x = -1\]

\[e. \quad 64^x – 8^{x+1} + 16 = 0\] \[\quad \quad \left(8^x\right)^2 – 8.(8^x) + 16 = 0 \quad y = 8^x\] \[\quad \quad y^2 – 89 + 16 = 0\] \[\quad \quad \quad (y – 4)(y – 4)\] \[\quad \quad \quad y = 4\] \[\quad \quad \quad y = 4 \quad \Rightarrow \quad 8^x = 4 \] \[\quad \quad 2^{3x} = 2^2 \quad \quad x = \frac{2}{3}\]

\[f. \quad 3^{2x+1} + 3^{x+2} = 3\frac{1}{3}\] \[\quad \quad 3.\left(3^x\right)^2 + 9.(3^x) = \frac{10}{3} \quad y = 3^x\] \[\quad \quad 3y^2 + 9y – \frac{10}{3} = 0 \quad (\div 3)\] \[\quad \quad y^2 + 3y – \frac{10}{9} = 0\] \[\quad \quad \quad (y + \frac{10}{3})(y – \frac{1}{3})\] \[\quad \quad \quad y = \frac{-10}{3} \lor y = \frac{1}{3}\] \[\quad \quad \quad y = \frac{-10}{3} \quad \Rightarrow \quad 3^x = \frac{-10}{3} \quad \Rightarrow \quad N/A\] \[\quad \quad \quad y = \frac{1}{3} \quad \Rightarrow \quad 3^x = \frac{1}{3} \quad \quad x = -1\]