zad 1 ^ [A] do potęgi A (a)/(b) - ułamek a przez b log(a)(b) - logarytm o podstawie a z liczby b a=36^[ log(6)(5) - (1)/(4) ] = 36^[log(6)(5)] : 36^[(1)/(4)] = = 6^[2log(6)(5)] : 6^[(1)(2)] = = 6^[log(6)(5^[2] )] : 6^[(1)(2)] = = 6^[log(6)(25)] : 6^[(1)(2)] = 25 : pierwiastek z 6 = (25pierw z 6):6 c=3^[2-log(3)(4)] = 3^[2] : 3^[log(3)(4)] = 9 : 4 = 2,25 zad 2 5^[log(5)(4)] + 10^[log( )(3)] = 4 + 3 = 7 zad 3 a=log( )(96^[0,25]) - 0,25log( )( 2/27 ) = = log( )(96^[0,25]) - log( )( (2/27)^0,25 ) = = log( )( (96 : 2/27)^0,25 ) = log( )(1296^0,25) = log( )(6) b = 1 / [log(6)(10)] = log(10)(6) = log( )(6) a=b zad 4 log(2)(x)=-2/3 2^[-2/3] = x x = 0,5^[2/3] x= 0,25^[1/3] x=pierwiastek stopnia 3 z 0,25
