Odpowiedzi

2010-01-21T23:13:11+01:00
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
1 5 1