Question 625447
I assume that you meant {{{16^(2k)= 4^(k-1)}}}.
Since {{{16=4^2}}}, we can write it as
{{{(4^2)^(2k)= 4^(k-1)}}}
Using the properties of powers, we can transform that equation into a friendlier one.
{{{(4^2)^(2k)= 4^(k-1)}}} --> {{{4^(2*2*k)= 4^(k-1)}}} --> {{{4^(4k)= 4^(k-1)}}}
Since {{{4^k}}} cannot be zero, we can divide by {{{4^k}}} both sides of the equal sign to get
{{{4^(4k)/4^k= 4^(k-1)/4^k}}}
Now we keep using properties of powers
{{{4^(4k)/4^k= 4^(k-1)/4^k}}} --> {{{4^(4k-k)=4^(k-1-k)}}} --> {{{4^(3k)=4^(-1)}}} --> {{{3k=-1}}} --> {{{highlight(k=-1/3)}}}
 
I understand that your other equation is {{{(x^(1/4)-6 )(x^(1/2)-3) = 0}}} or {{{(x^0.25-6 )(x^0.5-3) = 0}}} 
(I cannot get those nasty fractional exponents to show themselves nicely).
For that equation , I would just start by doing a change of variable, to get rid of those nasty fractional exponents.
I can define {{{y=x^(1/4)=x^0.25}}} <--> {{{x=y^4}}} and then {{{y^2=x^(1/2)=x^0.5}}}
I re-write the equation as
{{{(y-6)(y^2-3)=0}}}
I cannot use the solution {{{y=-sqrt(3)}}} because {{{y=x^(1/4)=x^0.25>=0}}} 
The other two y solutions give me good x solutions
{{{y=6}}} --> {{{x=6^4}}} --> {{{x=1296}}} and
{{{y=sqrt(3)}}} --> {{{x=(sqrt(3))^4}}} --> {{{x=9}}}