Maths questions

(log x/log 2) - (log y/log 2) + 3 = 0 ------(1)
and
(log y/log 2) = (log x/log 2)² - 2 (log x/log 2) - 15 ------(2)

Let u = log x/log 2 and v = log y/log 2. Then the question becomes,
u - v + 3 = 0 ------(1)
v = u² - 2u - 15 ------(2)

From (1),
v = u + 3
Substitute into (2),
u + 3 = u² - 2u - 15
u² - 3u - 18 = 0
(u - 6)(u + 3) = 0
u = 6 or u = -3
v = 9 or v = 0
i.e. ( u = 6 and v = 9 )　or　( u = -3 and v = 0 )

Changing back into x and y,
( log x/log 2 = 6 and log y/log 2 = 9 )　or　( log x/log 2 = -3 and log y/log 2 = 0 )
( log x = 6 log 2 and log y = 9 log 2 )　or　( log x = -3 log 2 and log y = 0 )
( log x = log 26 and log y = log 29 )　or　( log x = log 2-3 and log y = log 1 )
( x = 64 and y = 512 )　or　(x = 1/8 and y = 1)


(2)
(x-2)/2 < (3x+4)/3 < (6x-4)/6
3(x-2) < 2(3x+4) < (6x-4)【Multiply the whole inequality by 6】
3x-6 < 6x+8 < 6x-4

Since it is impossible for 6x+8 to be less than 6x-4, this question has no solution.


Hope it helps! ^^
Wish you a Happy 2007!

1)

( log x / log 2 ) - ( log y / log 2 ) + 3 = 0
and
(log y / log 2) = ( log x / log 2)^2 - 2(log x / log 2) - 15


(log x / log 2) - ( log y / log 2 ) + 3 = 0..............................(1)
(log y / log 2) = ( log x / log 2)^2 - 2(log x / log 2) - 15.........(2)

Put (2) into (1):
(log x / log 2) - [( log x / log 2)^2 - 2(log x / log 2) - 15] + 3 = 0
(log x / log 2) - ( log x / log 2)^2 + 2(log x / log 2) + 15 + 3 = 0
- ( log x / log 2)^2 + 3(log x / log 2) + 12 = 0
( log x / log 2)^2 - 3(log x / log 2) - 18 = 0
[( log x / log 2) - 6][( log x / log 2) + 3] = 0

( log x / log 2) - 6 = 0 or
( log x / log 2) + 3 = 0

log x / log 2 = 6 or
log x / log 2 = -3

log x = 6 log 2 or
log x = -3 log 2

log x = log 2^6 or
log x = log 2^(-3)

x = 2^6
or x = 2^(-3)

Hence x = 64 or x = 0.125
Put x = 64 into (2)
(log y / log 2)
= ( log 64 / log 2)^2 - 2(log 64 / log 2) - 15
= 6^2 - 2(6) - 15
= 9
log y = 9 log 2
log y = log 2^9
y = 2^9 = 512

Put x = 0.125 into (2)
(log y / log 2)
= ( log 0.125 / log 2)^2 - 2(log 0.125 / log 2) - 15
= (-3)^2 - 2(-3) - 15
= 0

log y = 0
y = 1


Hence, when x = 64, y = 512; x = 0.125, y = 1



2) (x-2)/2 &lt; (3x+4)/3 &lt; (6x-4)/6

(x-2)/2 &lt; (3x+4)/3..............x 6
3(x - 2) &lt; 2(3x + 4)
3x - 6 &lt; 6x + 8
3x &gt; -14
x &gt; -14/3

(3x+4)/3 &lt; (6x-4)/6...............x 6
2(3x+4) &lt; 6x -4
6x + 8 &lt; 6x - 4...................no root

Hence x &gt; -14/3

2006-12-31 17:10:15 補充：
Hence no root for this question.

2007-01-01 23:34:33 補充：
CHECK:For x = 64, y = 512(log64/log2)-(log512/log2) 3 = 6-9 3 = 0and(log512/log2) = 9(log64/log2)^2 - 2(log64/log2)-15 = 36-12-15 = 9For x = 0.125, y = 1(log0.125/log2)-(log1/log2) 3 = -3-0 3 = 0and(log1/log2) = 0(log0.125/log2)^2 - 2(log0.125/log2)-15 = 9 6-15 = 0