oblicz wiedząc, ze log_3x=-1/4
zalozenie: x>0log3(3^3*x^-2/3)= bo x^1/3:x to odejmujesz potegi=log3 (3^3)+log3 (x^-2/3)=3-(2/3)log3 x=3-(2/3)*(-1/4)=3+(2/12)=3+1/6=3 i 1/6
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zalozenie: x>0
log3(3^3*x^-2/3)= bo x^1/3:x to odejmujesz potegi
=log3 (3^3)+log3 (x^-2/3)=3-(2/3)log3 x=3-(2/3)*(-1/4)=3+(2/12)=3+1/6=3 i 1/6