(square root y-3)^2=(1-square root 2y-4)^2. how do i do this???? i am clueless, pleaseeee help!!!!

Question

i^-7. how do I do this please describe! thanks for your time and help, I really appreciate it:)

Answer 1

i is -1 so
i^2 = +1
i^3 = -1 = i
i^4 = +1
Note that when the exponent is odd the answer is always -1 or i.
When the exponent is even the answer is always +1, so
i^-7 = (1/i^7) = 1/i = 1/-1 = -1 = i.
Check my work.

Answer 2

I will answer your first question.
looks like you were solving

√(y-3) = 1 - √(2y-4) and you are squaring both sides. You will get

y-3 = 1 - 2√(2y-4) + 2y - 4
2√(2y-4) = y
let's square again

4(2y-4) = y^2
y^2 - 8y + 16 = 0
(y-4)^2 = 0
y-4 = 0
y = 4

since we squared, our answer must be verified in the original equation

LS = √y-3 = √(4-3) = 1
RS = 1 - √(2y-4) = 1 - 2 = -1

So there is no solution.

Answer 3

thanks that makes alot of sense:)