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=?

2 Answers

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  • 9 years ago
    Favorite Answer

    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
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