Maths

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Q 1 : 3^2x-3^x-6=0

A 1 :

3^2x-3^x-6=0
Substitute u = e^(x^2) into the left hand side:
u^2-u-6 = 0
Add 6 to both sides:
u^2-u = 6
Add 1/4 to both sides:
u^2-u+1/4 = 25/4
Factor the left hand side:
(u-1/2)^2 = 25/4
Take the square root of both sides:
|(u-1/2)| = 5/2
Eliminate the absolute value:
u-1/2 = -5/2 or u-1/2 = 5/2
Add 1/2 to both sides:
u = -2 or u-1/2 = 5/2
Substitute back for u = 3^x:
3^x = -2 or u-1/2 = 5/2
Take the logarithm to the base 3 of both sides:
x = (iπ+log(2))/(log(3)) or u-1/2 = 5/2
Add 1/2 to both sides:
x = (iπ+log(2))/(log(3)) or u = 3
Substitute back for u = 3^x:
x = (iπ+log(2))/(log(3)) or 3^x = 3
Take the logarithm to the base 3 of both sides:
x = (iπ+log(2))/(log(3)) or x = 1

Answer : x = 1 or (iπ+log(2))/(log(3))

Q 2 : 3(3^2x0-10(3^x)=3=0

A 2 :

What do you mean of ' =3=0 ' and ' 3^2x0 ' ?

Q 3 : (logx)^2-logx-12=0

A 3 :

(logx)^2-logx-12=0

Substitute u = log(x):
u^2-u-12 = 0
Factor the left hand side:
(u+3)(u-4)=0
u = -3 or 4
Substitute back for u = log(x):
log(x) = -3 or log(x) = 4
Cancel logarithms by taking exp of both sides:
x = 1/e^3 or x = e^4
Now test that these solutions are appropriate by substitution into the original equation:
Check the solution x = 1/e^3:
log^2(x)-log(x)-12 => -12-log(1/e^3)+log^2(1/e^3) = 0
So the solution is correct.
Check the solution x = e^4:
log^2(x)-log(x)-12 => -12-log(e^4)+log^2(e^4) = 0
So the solution is correct.

Answer : x = 1/e^3 or e^4 .

Q 4 : 2x-7√x+6=0

A 4 :

2x-7√x+6=0

Subtract 2x+6 from both sides:
-7√x = -2x-6
Divide both sides by -7:
√x = 1/7 (2x+6)
Square both sides:
x = [(2x+6)^2]/49
Expand out terms on the right hand side:
x = (4 x^2)/49+(24 x)/49+36/49
Subtract ((4 x^2)/49+(24 x)/49+36/49) from both sides:
-(4x^2)/49+(25 x)/49-36/49 = 0
49[-(4x^2)/49+(25 x)/49-36/49] = 0
-4x^2 +25x-36 = 0
-(-4x^2+25x-36)=0
4x^2-25x+36=0
(4x-9)(x-4)=0
4x=9 or x=4
x= 9/4 or 4

Answer : x = 9/4 or 4

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由於字數限制，Q5 , Q6 及 Q7 將到意見欄回答，不便之處，敬請原諒。

2011-07-21 16:38:45 補充：
Q 5 : x-4√x=-3

A 5 :

x - 4√x = -3
Subtract x from both sides:
-4√x = -x-3
Divide both sides by -4:
√x = (x+3)/4
Square both sides:
x = [(x+3)^2]/16
Expand out terms on the right hand side:
x = x^2/16+(3x)/8+9/16

2011-07-21 16:38:57 補充：
Subtract (x^2/16+(3x)/8+9/16) from both sides:
-x^2/16+(5x)/8-9/16=0
16[-x^2/16+(5x)/8-9/16]=0
-x^2+10x-9=0
-(-x^2+10x-9)=0
x^2-10x+9=0
(x-1)(x-9)=0
x=1or9

Answer : x = 1 or 9 .

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2011-07-21 16:39:05 補充：
Q 6 : i^18=?

A 6 :

i^18
= i^16 × i^2
= (i^4)^4 × ( i × i )
= (i×i×i×i)^4 × (-1)
= (-1×-1)^4 × (-1)
= 1^4 × (-1)
= 1 × (-1)
= -1

Answer : i^18 = -1 .

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2011-07-21 16:39:31 補充：
Q 7 : i^43=?

A 7 :

i^43
= i^40 × i^3
= (i^4)^10 × (i×i×i)
= (i×i×i×i)^10 × (-1×i)
= - [ (-1×-1)^10 × 1 × i ]
= - ( 1^10 × i )
= - ( 1 × i )
= -i

Answer : i^43 = -i .

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Put u = 3^x
3^2x - 3^x - 6 = 0
(3^x)^2 - (3^x) - 6 = 0
u^2 - u - 6 = 0
(u - 3)(u + 2) = 0
u = 3 or u = -2
3^x = 3 or 3^x = -2 (rejected)
3^x = 3^1
x = 1


=====
Put u = 3^x
3(3^2x) - 10(3^x) + 3 = 0
3(3^x)^2- 10(3^x) + 3 = 0
3u^2- 10u + 3 = 0
(u - 3)(3u - 1) = 0
u = 3 or u = 1/3
3^x = 3 or 3^x = 1/3
3^x = 3^1 or 3^x = 3^-1
x = 1 or x = -1


=====
Put u = logx
(logx)^2 - logx - 12 = 0
u^2- u - 12 = 0
(u - 4)(u + 3) = 0
u = 4 or u = -3
logx = 4 or logx = -3
x = 10^4 or x = 10^-3
x = 10000 or x = 1/1000


=====
Put u = √x
2x - 7√x + 6 = 0
2u^2- 7u + 6 = 0
(2u - 3)(u - 2) = 0
u = 3/2 or u = 2
√x = 3/2 or √x = 2
x = 9/4 or x = 4


=====
Put u = √x
x - 4√x = -3
u² - 4u + 3 = 0
(u - 1)(u - 3) = 0
u = 1 or u = 3
√x = 1 or √x = 3
x = 1 or x = 9


=====
i = √-1
i^2 = (√-1)^2 = -1
i^3 = i^2*i = -i
i^4 = (i^2)^2 = (-1)^2 = 1

i^18
= (i^16)*(i^2)
= (i^4)^4 * (i^2)
= (1)^4 * (-1)
= -1


====
i =√-1
i^2 = (√-1)^2 = -1
i^3 = i^2*i = -i
i^4 = (i^2)^2 = (-1)^2 = 1

i^43
= (i^40) * (i^3)
= (i^4)^10 * (i^3)
= (1)^10 * (-i)
= -i