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Math Help - 4(2^2x) + 31 (2^x) - 8 = 0.... please help.

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    4(2^2x) + 31 (2^x) - 8 = 0.... please help.

    Hi, I am not sure if I am in the correct place but here I go....


    I am taking Advance Functions, and I am currently in a unit about Exponential and Logarithmic Equations.

    I have to solve 4(22x) + 31 (2x) - 8 = 0

    this is what I have done so far...

    4(22x) + 31 (2x) - 8 = 0
    4((2x)2) + 31 (2x) - 8 = 0 (the book says " you can do a substitution here to make things a little easier.)

    4z2 +31z- 8 = 0 ( by replacing 2x with Z) ***I am stuck here***






    Here is a similar example the book has... (i used it to help guide me through this one, but i couldn't get pass the 31z.)

    3(32x) - 28(3x) + 9 = 0

    3 ((3x)2) - 28 (3x) + 9 = 0

    3z2 - 28z + 9 = 0

    3z2- 27z - z + 9 = 0

    3z(z-9) - (z-9) = 0

    (3z -1) (z-9) = 0 (by factoring)

    3z -1 = 0 or z - 9 = 0

    z= 1/3 or z= 9

    3x = 1/3 or 3x= 9 (by replacing z with 3x)

    Therefore, x = -1 or x = 2.



    Anyways I really do hope someone can help with this... I've been stuck on this problem for a couple days now & i just want to be finish!

    THANKS IN ADVANCED!
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    Re: 4(2^2x) + 31 (2^x) - 8 = 0.... please help.

    For the first problem, I see a quadratic problem that you can solve either by factoring or using the quadratic formula!
    Ill give this a go:

    4z^2 +31z- 8 = 0

    this can be factored. I know right away because I look at the discriminant which when solved yields a perfect square of 1089.

    So,

    4z^2 +31z- 8 = 0
    (4z -1)(1z +8 )=0

    There for z = 1/4 and -8


    I only took grade 11 functions, but just looking at your problem, I hope what I did helps!
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    Re: 4(2^2x) + 31 (2^x) - 8 = 0.... please help.

    Quote Originally Posted by marrzbarrz View Post
    I am taking Advance Functions, and I am currently in a unit about Exponential and Logarithmic Equations.
    I have to solve 4(22x) + 31 (2x) - 8 = 0
    this is what I have done so far...
    4(22x) + 31 (2x) - 8 = 0
    4((2x)2) + 31 (2x) - 8 = 0 (the book says " you can do a substitution here to make things a little easier.)

    4z2 +31z- 8 = 0 ( by replacing 2x with Z) ***I am stuck here***

    4z^2+31z-8=(4z-1)(z+8)=0 so z=~?~\&~z=~?
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