SOLUTION: Please help me solve this equation 16^2k= 4^k-1 and (x^1/4-6 )(x^1/2-3) = 0

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Question 625447: Please help me solve this equation
16^2k= 4^k-1

and (x^1/4-6 )(x^1/2-3) = 0
Answer by KMST(5328) About Me  (Show Source):
You can put this solution on YOUR website!
I assume that you meant 16%5E%282k%29=+4%5E%28k-1%29.
Since 16=4%5E2, we can write it as
%284%5E2%29%5E%282k%29=+4%5E%28k-1%29
Using the properties of powers, we can transform that equation into a friendlier one.
%284%5E2%29%5E%282k%29=+4%5E%28k-1%29 --> 4%5E%282%2A2%2Ak%29=+4%5E%28k-1%29 --> 4%5E%284k%29=+4%5E%28k-1%29
Since 4%5Ek cannot be zero, we can divide by 4%5Ek both sides of the equal sign to get
4%5E%284k%29%2F4%5Ek=+4%5E%28k-1%29%2F4%5Ek
Now we keep using properties of powers
4%5E%284k%29%2F4%5Ek=+4%5E%28k-1%29%2F4%5Ek --> 4%5E%284k-k%29=4%5E%28k-1-k%29 --> 4%5E%283k%29=4%5E%28-1%29 --> 3k=-1 --> highlight%28k=-1%2F3%29

I understand that your other equation is %28x%5E%281%2F4%29-6+%29%28x%5E%281%2F2%29-3%29+=+0 or %28x%5E0.25-6+%29%28x%5E0.5-3%29+=+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%5E%281%2F4%29=x%5E0.25 <--> x=y%5E4 and then y%5E2=x%5E%281%2F2%29=x%5E0.5
I re-write the equation as
%28y-6%29%28y%5E2-3%29=0
I cannot use the solution y=-sqrt%283%29 because y=x%5E%281%2F4%29=x%5E0.25%3E=0
The other two y solutions give me good x solutions
y=6 --> x=6%5E4 --> x=1296 and
y=sqrt%283%29 --> x=%28sqrt%283%29%29%5E4 --> x=9