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    81^x + 81 = 30(9^x)

    Find the value of x.
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    :confused:
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    (Original post by The Chameleon)
    81^x + 81 = 30(9^x)

    Find the value of x.

    can you confirm the 30 in the question, please?
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    substitute for u=9^x
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    81^x = (9^2)^x = 9^2x
    u=9^x
    u^2 + 81 = 30u
    So.. solve u^2 - 30u + 81 = 0
    (u + 27)(u + 3) = 0
    u = -27 or -3

    Then substitute the u back.
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    (Original post by Fermat)
    substitute for u=9^x
    81^x + 81 = 30(9^x)

    Like Fermat said. Make it into a quadratic with u=9^x

    so (9^x)^2+81-30(9^x)=0
    u^2-30u+81=0
    (u-15)^2+81-225=0
    so (u-15)^2=144
    u-15=±12
    so u=3 or 27

    so either 9^x=3,
    3^(2x)=3
    so 2x=1, so x=1/2

    or 9^x=27
    so 3^(2x)=3^3
    so 2x=3, so x=3/2
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    Ok... ignore mine
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    my teacher taught me this method.. i'm not sure if it's right or not:

    81^x + 81 = 30(9^x)

    (3^4)^x + (3^4)^1 = 30(3^2)^x
    as 3 is a common, you can substitute y in for it instead
    so you get y^4x + y = 30y^2
    y^4x - 30y^2x + y = 0
    something is wrong in my arithmetic there, but that's the method

    4x+ 4 = 60x
    4x=56
    x=14
    [also answer may be wrong, but method of substitution is right. hope that helps
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    (Original post by The Chameleon)
    81^x + 81 = 30(9^x)

    Find the value of x.
    81^x + 81 = 30(9^x)
    (9^x)^2 - 30*(9^x) + 81 = 0
    (9^x - 27)(9^x - 3) = 0
    9^x = 27 or 9^x = 3
    x = 3/2 or 1/2

    yeah others got the same Im glad to see
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    (Original post by mik1a)
    81^x + 81 = 30(9^x)
    (9^x)^2 - 30*(9^x) + 81 = 0
    (9^x - 27)(9^x - 3) = 0
    9^x = 27 or 9^x = 3
    x = 3/2 or 1/2

    yeah others got the same Im glad to see
    Can someone explain this to me?

    Wouldn't (9^x)^2 give 81^2x whereas we are given 81^x?
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    (Original post by The Chameleon)
    Can someone explain this to me?

    Wouldn't (9^x)^2 give 81^2x whereas we are given 81^x?
    no

    (9^x)^2=9^x.9^x
    =9^(2x)
    =(9^2)^x
    =81^x
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    81 is the same as 9^2

    => 81^x = (9^2)^x.

    When indicies are like that we multiply them together

    => (9^2)^x = 9^2x

    However, it doesn't matter which way round we multiply them, it will still yield the same answer. So,

    => (9^2)^x = (9^x)^2

    This second form is more useful as it allows you to substitue in for u=9^x on both sides of the equation.
 
 
 
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