Wyznacz x : a) log przy podstawie x 32=-5 b) log przy podstawie x 0,125=0,75

[latex]x > 0 wedge x eq 1 Rightarrow xin (0,1)cup (1,infty)[/latex] a)   [latex]log_x32 = -5\ x^{-5} = 32\ x^{-5} = 2^5\ x^{-5} = (frac{1}{2})^{-5}\ x=frac{1}{2}[/latex]   b)   [latex]log_x0,125 = 0,75\ x^{0,75} = 0,125\ x^{frac{3}{4}} = 0,5^3 \ x=(0,5^3)^{frac{4}{3}}\ x=0,5^4 = 0,0625[/latex]  

a) log x [ 32] = - 5  <=> x^(-5) = 32 <=> x = 1/2 b) log x [ 0,125 ] = 0.75 <=> log x [ 1/8] = 3/4  <=> x^(3/4) = 1/8 <=> x = [ 1/8]^(4/3) = (1/8)*(1/2) = 1/16