Soal Eksponen & Logaritma, tolong dibantu kak/bang​

Jawaban:@Ara1412 1. (9^(x+4))^(1/2) - log 81^(x-5) = 0, nilai x = ? - (9^(x+4))^(1/2) - log 81^(x-5) = 0- 3^(x+4) - (x-5) log 81 = 0 (Asumsi: log adalah logaritma basis 3)- 3^(x+4) - (x-5) * 4 = 0- 3^(x+4) = 4(x-5)- 3^(x+4) = 4x - 20- Dengan mencoba beberapa nilai x, kita temukan bahwa x = -2 memenuhi persamaan:- 3^(-2+4) = 4(-2) - 20- 3^2 = -8 - 20- 9 = -28 (Tidak memenuhi)- Karena tidak ada solusi sederhana, kita perlu menggunakan metode numerik atau grafik untuk menemukan solusi yang tepat. Namun, tanpa alat bantu, kita tidak dapat memberikan solusi eksak. 2. (2^(1/2) + 2^(1/3)) / (2^(-1/2) + 2^(-1/3)) = 4^x, nilai x = ? - (2^(1/2) + 2^(1/3)) / (2^(-1/2) + 2^(-1/3)) = 4^x- (2^(1/2) + 2^(1/3)) / (1/2^(1/2) + 1/2^(1/3)) = 4^x- (2^(1/2) + 2^(1/3)) / ((2^(1/3) + 2^(1/2)) / (2^(1/2) * 2^(1/3))) = 4^x- (2^(1/2) + 2^(1/3)) * (2^(1/2) * 2^(1/3)) / (2^(1/3) + 2^(1/2)) = 4^x- 2^(1/2) * 2^(1/3) = 4^x- 2^(1/2 + 1/3) = 2^(2x)- 2^(5/6) = 2^(2x)- 5/6 = 2x- x = 5/12 3. (3x^(-1) - y^(-2)) / (x^(-2) + 2y^(-1)) = ? - (3x^(-1) - y^(-2)) / (x^(-2) + 2y^(-1))- (3/x - 1/y^2) / (1/x^2 + 2/y)- ((3y^2 - x) / (xy^2)) / ((y + 2x^2) / (x^2y))- ((3y^2 - x) / (xy^2)) * ((x^2y) / (y + 2x^2))- (x(3y^2 - x)) / (y(y + 2x^2))- (3xy^2 - x^2) / (y^2 + 2x^2y) Kesimpulan: 1. Nilai x pada persamaan pertama memerlukan metode numerik atau grafik untuk solusi eksak.2. Nilai x pada persamaan kedua adalah 5/12.3. Bentuk sederhana dari persamaan ketiga adalah (3xy^2 - x^2) / (y^2 + 2x^2y).Penjelasan dengan langkah-langkah:Tentu, mari kita selesaikan soal-soal dari gambar tersebut: 4. (3a²b^(1/3)c^(-1/2) / ab^(-1)c^(-2))^(-1) - (3a²b^(1/3)c^(-1/2) / ab^(-1)c^(-2))^(-1)- (3a^(2-1)b^(1/3 - (-1))c^(-1/2 - (-2)))^(-1)- (3ab^(4/3)c^(3/2))^(-1)- (1 / (3ab^(4/3)c^(3/2)))- 1 / (3a * b^(4/3) * c^(3/2)) 5. 2^(x+5) + 4^(x+1) = 512 - 2^(x+5) + 4^(x+1) = 512- 2^(x+5) + (2^2)^(x+1) = 512- 2^(x+5) + 2^(2x+2) = 512- 2^(x+5) + 2^(2x+2) = 2^9- Misalkan x = 2:- 2^(2+5) + 2^(2(2)+2) = 2^7 + 2^6 = 128 + 64 = 192 (Tidak memenuhi)- Misalkan x = 3:- 2^(3+5) + 2^(2(3)+2) = 2^8 + 2^8 = 256 + 256 = 512 (Memenuhi)- Jadi, x = 3 6. p = (x^(3/2) + x^(1/2))(x^(1/3) - x^(-1/3)) dan q = (x^(1/2) + x^(-1/2))(x - x^(1/3)), maka nilai p/q = ? - p = (x^(3/2) + x^(1/2))(x^(1/3) - x^(-1/3))- p = x^(1/2)(x + 1) * x^(-1/3)(x^(2/3) - 1)- p = x^(1/2 - 1/3)(x + 1)(x^(2/3) - 1)- p = x^(1/6)(x + 1)(x^(2/3) - 1)- q = (x^(1/2) + x^(-1/2))(x - x^(1/3))- q = x^(-1/2)(x + 1) * x^(1/3)(x^(2/3) - 1)- q = x^(-1/2 + 1/3)(x + 1)(x^(2/3) - 1)- q = x^(-1/6)(x + 1)(x^(2/3) - 1)- p/q = (x^(1/6)(x + 1)(x^(2/3) - 1)) / (x^(-1/6)(x + 1)(x^(2/3) - 1))- p/q = x^(1/6) / x^(-1/6)- p/q = x^(1/6 + 1/6)- p/q = x^(1/3) Kesimpulan: 4. 1 / (3a * b^(4/3) * c^(3/2))5. x = 36. p/q = x^(1/3)

Answered by ara1412 | 2025-08-10