Cans someone solve this logarithmic question for me?

log2 y + (1/2)log2 y^6 = 8
log2 y + log2 y^(6/2) = 8
log2 y + log2 y^3 = 8
log2 (y * y^3) = 8
log2 y^4 = 8
4 log2 y = 8
log2 y = 8/4
log2 y = 2
y = 2^2 = 4
Ans: 4
_________________.

The correct answer is y=4.

Why: You can simplify this to:
log(base2) of y + log(base2) of (y^(6))^(1/2) = log(base 2) 2^8 (you can raise up the 1/2 into the y^6 part of the log also rewrote right side)
log(base 2) y + log(base2) of y^3 = log (base 2) 2^8 (simplifying sqrt(y^6)=y^3)
log(base 2) y^4 = log(base 2) 2^8
y^4=2^8
y=2^2=4

You can plug in y=4 and see that it in fact works. Plugging in the incorrect solution of y=1 yields: 0+0=8, which is obviously incorrect.

Thus, the answer is y=4

[Note: I sort of lied, sqrt(y^6)= absolute value of y^3, but it wasn't relevant here and I didn't want to add extra confusion. This is just a disclaimer and if it confused you ignore it!}

Method: collect all possible roots and discard those that don't satisfy the equation (this is a good example of why math teachers shouldn't profess that log(x²) = 2 log(x) is an identity without discussing its limitations; for example, consider x = -1):
log2(y) + 1/2 log2(y^6) = 8
log2(y^8) = 16
y^8 = 65536
y = 4 e^(ki?/4) {k = 0,1,2,3,4,5,6,7}
discard k = 2, 3, 4, 5  (they do not satisfy the equation).
or in standard form,
y = 4, ?2(2+2i), -4i, ?2(2-2i),

Answer: y = 4, 2?2(1 + i), -4i, 2?2(1 - i)