How would I factor 4^2x + 2(4^x) - 3 = 0?

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  • 4 weeks ago

    4^(2x)+2(4^x)-3=0

    =>

    (4^x)^2+2(4^x)-3=0

    =>

    [(4^x)+3][(4^x)-1]=0

    or

    x[x-ln(-3)/ln(4)]=0

  • 4 weeks ago

     4^2x + 2(4^x) - 3 = 0

     (4^)^2 + 2(4^x) - 3 =  0 

    Think of 4^x = y 

    Substitute 

    y^2 + 2y - 3 = 0 

    Factor 

    ( y + 3)(y - 1) = 0 

    Hence 

    4^x = -3  (Unresolved)

    4^x = 1 

    x = 0 

  • 4 weeks ago

    4^(2x) + [2 * 4^(x)] - 3 = 0

    4^(x + x) + [2 * 4^(x)] - 3 = 0

    [4^(x) * 4^(x)] + [2 * 4^(x)] - 3 = 0 → let: a = 4^(x) → where: a > 0 because exponential

    [a * a] + [2 * a] - 3 = 0

    a² + 2a - 3 = 0

    a² + 2a = 3

    a² + 2a + 1 = 4

    (a + 1)² = 4

    a + 1 = ± 2

    a = - 1 ± 2 → recall the condition: a > 0

    a = - 1 + 2

    a = 1 → recall: a = 4^(x)

    4^(x) = 1

    x = 0

  • 4 weeks ago

    (4^2)x + 2(4^x) - 3 = 0

    Krishnamurthy gave the correct solution which uses the Product Log (Lambert) Function.

    Or did you mean:

    4^(2x) + 2(4^x) - 3 = 0

    Let u = 4^x, so u^2 = 4^(2x), so we have:

    u^2 + 2u - 3 = 0

    u^2 + 3u - u - 3 = 0

    u(u + 3) - (u + 3) = 0

    (u - 1)(u + 3) = 0

    u - 1 = 0 or u + 3 = 0

    4^x = 1 or 4^x = -3

    If 4^x = 1 then:

    x = (ln(1) + 2*i*pi*n) / ln(4), for any integer n

    x = (i*pi*n) / ln(2), for any integer n

    x =~ 4.5323601418271938096276829457167in, for any integer n

    If x is a real number, then n = 0, so x = 0

    If 4^x = -3 then:

    x = (ln(-3) + 2*i*pi*n) / ln(4), for any integer n

    x = (ln(-3) + i*pi*n) / ln(4), for any even number n

    x = (ln(3) + i*pi*n) / ln(4), for any odd number n

    x =~ 0.79248125036057809072686947197391 + 2.2661800709135969048138414728583n, for any odd number n

    https://www.wolframalpha.com/input/?i=4%5E%282x%29...

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  • 4 weeks ago

    4^2x + 2(4^x) - 3 = 0

    2 (8 x + 4^x) = 3

    x ≈ 0.0529747

  • rotchm
    Lv 7
    1 month ago

    The idea here is to notice that your equation is a quadratic. Do you see it?

    And since you know how to solve a quadratic, you are essentially done. 

    Hopefully no one will spoil you the answer thereby depriving you from your personal enhancement; that would be very inconsiderate of them. 

  • 1 month ago

    Assuming 4^2x should be 4^(2x),

    4^(2x) + 2(4^x) - 3 = 0

    Let u = 4^x.

    u^2 - 2u - 3 = 0

    (u - 3)(u + 1) = 0

    u = 3 or u = -1

    4^x = 3 or 4^x = -1

    4^x = -1 has no real solutions, leaving just

    4^x = 3

    x = log(base 4)3

  • 1 month ago

    This is an equation. If you wish to factorize the left side then proceed as follows :

    4^(2x) + 2 ( 4^x ) - 3 

    This can be re-written as  =  (4^x)²  + 2 ( 4^x ) - 3 

    Let 4^(x) = A then --

    =>  4^(2x) + 2 ( 4^x ) - 3  =  A² + 3 A - A - 3

    =>  A ( A + 3 ) - 1 ( A + 3 ) 

    => ( A - 1 ) ( A + 3 )

    Putting  A  =  4^x

    4^(2x) + 2 ( 4^2 ) - 3  =  ( 4^x - 1 ) ( 4^x + 3 )  ................... Required factors.(Answer)

    If you want to calculate x from this equation, then put --

    ( 4^x - 1 ) ( 4^x + 3 )  =  0

    =>  ( 4^x - 1 )  =  0   ===>>>  4^x  =  1 ====>>>  x = 0 

    And  ( 4^x + 3 ) = 0  =>  4^x  =  - 3

    => 2^(2x)  =  - 3

    Power on the Left hand side is even, hence this equation will have two complex roots.

  • 1 month ago

    4²ˣ = (4ˣ)²

    so, (4ˣ)² + 2(4ˣ) - 3 = 0

    If y = 4ˣ then,

    y² + 2y - 3 = 0

    i.e. (y + 3)(y - 1) = 0

    Hence, y = -3 or y = 1

    so, 4ˣ = 1 or 4ˣ = -3...not possible

    Then, x = 0 is the only solution

    Checking gives: 4⁰ + 2(4⁰) - 3 => 1 + 2 - 3 = 0

    :)>  

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