myteapot172000 asked in 科學及數學數學 · 9 years ago

# Maths

3^2x-3^x-6=0

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

(logx)^2-logx-12=0

2x-7√x+6=0

x-4√x=-3

i^18=?

i^43=?

Rating
• 9 years ago

Hi ! I am lop****** , feel happy to answer your question.

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

======================================

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 .

======================================

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 .

======================================

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 .

======================================

Source(s): Hope I Can Help You ^_^ ( From website + Me )
• 土扁
Lv 7
9 years ago

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

Source(s): 土扁