Question 189805
Are you sure that the 0.3 isn't 0.3333 (where the 3's repeat)? If the 3's repeat, then {{{1/3=0.3333}}} (where the 3's repeat)




{{{log(9,(0.3333))}}} Start with the given expression



{{{log(9,(1/3))}}} Replace {{{0.3333}}} with {{{1/3}}}



{{{y=log(9,(1/3))}}} Set the expression equal to "y"



{{{9^y=1/3}}} Use the property stated above to rewrite the equation



{{{(3^2)^y=3^(-1)}}} Rewrite {{{9}}} as {{{3^2}}} and {{{1/3}}} as {{{3^(-1)}}}



{{{3^(2y)=3^(-1)}}} Multiply the exponents



{{{2y=-1}}} Since the bases are equal, the exponents are equal.



{{{y=-1/2}}} Divide both sides by 2 to isolate "y"



Since {{{y=log(9,(1/3))}}}, this means that {{{log(9,(1/3))=-1/2}}} and {{{log(9,(0.3333))=-1/2}}} (where the 3's repeat)