Question

1

Answer 1

\[\huge \log_{2}x-\log_{4}y=4\]

\[\huge \log_{2}(x-2y)=5\]

Answer 2

log(2) x - log(4) y = 4
log(2) x - [log(2) y]/[log(2) 4] = 4
log(2) x - [log(2) y]/2 = 4
[2 log(2) x]/2 - [log(2) y]/2 = 4
log(2) x^2 - log(2) y = 8
log(2) x^2/y = 8
x^2/y = 2^8________________(1)

log(2) x-2y = 5
x - 2y = 2^5
x = 2y + 32________________(2)

substitute equation (2) into equation (1);
(2y + 32)^2/y = 2^8
(2y + 32)^2 = 256y
4y^2 + 128y + 1024 = 256y
4y^2 -128y + 1024 = 0
y^2 - 32y + 256 = 0
(y - 16)(y - 16) = 0
y = 16

substitute y = 16 into equation (2);
x = 2(16) + 32
x = 64

Answer 3

2^5 = x - 2y
32 = x -2y

32 + 2y = x..that's a good start

Answer 4

@lgbasallote give medal

Answer 5

log_2 (32 + 2y) - log_4 (y) = 4

ahhh if only i remember how to turn log_4 to log_2 or vice versa -__-

Answer 6

hmmm depends...is rohan right @.sam.?

Answer 7

very dumb and easiest problem

Answer 8

@Rohangrr don't use offensive languages, and I don't understand what you're working

Answer 9

@.Sam. I'm really soooooooory well yes its the easiest one

Answer 10

well..the hint says it's a hard one so i guess it cant be solved simply..there must be something else here..hmmm.

Answer 11

@Igbasallote i agree to u