SOLUTION: If log₂x+ log₂y=4,show that a)log₄xy=2 b)log₈xy=4/3

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Question 1199386: If log₂x+ log₂y=4,show that
a)log₄xy=2
b)log₈xy=4/3

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
log2(x) + log2(y) = 4 becomes:
log2(xy) = 4 if and only if:
2^4 = xy

2^4 = 16, therefore:
16 = xy

4^2 = 16, therefore:
4^2 = xy if and only if:
log4(xy) = 2

8^(4/3) = 16, therefore:
8^(4/3) = xy if and only if:
log8(xy) = 4/3

you can use your calculator to confirm that these are true.

use the log base conversion formula to get loga(x) = logb(x) / logb(a).
with your calculator, you can convert all of these to base 10 which is the log function of your calculator.

since xy = 16, then:

log2(xy) = log2(16) = log(16) / log(2) = 4. confirming that log2(xy) = 4

log4(xy) = 2 = log4(16) = log(16) / log(4) = 2, confirming that log4(xy) = 2

log8(xy) = 4/3 = log8(16) = log(16) / log(8) = 1 and 1/3 = 4/3, confirming that log8(xy) = 4/3

this should be sufficient for what you need, but if not, let me know what else you need to show that selections a and b are true.