Tentukan nilai x yang memenuhi 2.3 log x - 3 log 2 = 3 log (4+x) !

[tex]2. {}^{3} log(x) - {}^{3} log(2) = {}^{3} log(4 + x) \: dgn \: syarat \: x \: lbh \: dr \: 0 \\ {}^{3} log(x {}^{2} ) - {}^{3} log(2) = {}^{3} log(4 + x) \\ {}^{3} log( \frac{x {}^{2} }{2} ) = {}^{3} log(4 + x) \\ \frac{x {}^{2} }{2} = 4 + x \\ x {}^{2} = 2(4 + x) \\ x {}^{2} - 2(4 + x) = 0 \\ x {}^{2} - 2x - 8 = 0 \\ \\ selesaikan \: dgn \: rumus \: abc \\ sehingga \\ x = \frac{2 + 6}{2} \\ = 4 \\ dan \\ x = \frac{2 - 6}{2} \\ = - 2 \: (tdk \: memenuhi) \\ \\ maka \\ x = 4[/tex]