tentukanlah harga x yg memenuhi persamaan a.3^5x-1 =27^x+3 b.akar 4^x+4 =(1/16)^5-x tolong di jawab

Kelas : X SMA
Mapel : Matematika Peminatan
Bab : Fungsi Eksponensial dan Logaritma
Indikator : Persamaan Eksponen dan Logaritma

Langkah Penyelesaian dan Jawaban :
[tex]a) \ 3^{5x-1} = 27^{x+3} \\ 3^{5x-1} = (3^3)^{x+3} \\ 5x - 1 = 3(x+3) \\ 5x-1 = 3x + 9 \\ 5x - 3x = 9 + 1 \\ 2x = 10 \\ x = 5 \\\\ b) \ \sqrt{4^{x+4}} = ( \frac{1}{16} )^{5-x} \\ (4^{x+4})^ \frac{1}{2} = (2^{-4})^{5-x} \\ 4^{ \frac{x + 4}{2} } = (2^{-4})^{5-x} \\ (2^2)^{\frac{x+4}{2}} = (2^{-4})^{5-x} \\ 2^{x+4} = (2^{-4})^{5-x} \\ x + 4 = - 4(5-x) \\ x+4 = - 20 + 4x \\ x - 4x = - 20 - 4 \\ - 3x = - 24 \\ x = 8[/tex]

[tex]a. \: 3 ^{5x - 1} = {27}^{x + 3} \\ {3}^{5x - 1} = ( {3}^{3(x + 3)}) \\ 5 - 1 = 3x + 9 \\ 5x - 3x = 10 \\ 2x = 10 \\ x = 5[/tex]