[tex]2\log4 + \log3 -\log2 + \log5= \\ \\ = \log 2^4+ \log3 -\log2 + \log5 = \\ \\ = \log 16+ \log3 -\log2 + \log5 = \\ \\ = \log\Big(16\cdot 3:2\cdot 5\Big) = \log\Big(\dfrac{16\cdot 3\cdot 5}{2}\Big) = \log(8\cdot 3\cdot 5) = \\ \\ =\log120[/tex]
