rozwiąż równania: [latex]sqrt{0,2}= 25^x[/latex]       [latex]frac{4}{3}=frac{8}{27}^x cdot 3[/latex]

1) Pierwiastek z 0,2 =  (1/5)^(1/2) zatem mamy (1/5)^(1/2) = [ 5^2]^x [ 5 ^(-1)]^(1/2) = 5 ^(2x) 5^( -1/2) = 5^(2x) -1/2 = 2x  / : 2 x = - 1/4 = - 0,25 ================ 2) 4/3 ={ [ 8/27]^x} * 3  / : 3 4/9 = [ 8/27]^x [2/3]^2 = [ (2/3)^3]^x [2/3]^2 = [2/3]^(3x) 2 = 3x / : 3 2/3 = x x = 2/3 ==========