Question 128840
I suspect you mean {{{log(x,(1/64))=-3}}}


Remember that {{{log(b,x)=y}}} means that {{{b^y=x}}}


So for your problem, the base is x, the exponent is -3, and the result of raising x to the -3 power is {{{1/64}}}.


{{{x^-3=1/64}}}


But remember that {{{a^(-n)=1/a^n}}}, so


{{{1/x^3=1/64}}}, and that means that 


{{{x^3=64}}}


And since the cube root of 64 is 4,


{{{x=4}}}