log(base2)x+log(base2)(x+5)=k?

solve for k in terms of k.

solve for x when if k=7

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  • 1 decade ago
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    "Solve for k in terms of k" doesn't make sense: k=k is obvious.

    I assume one of those k's ought to have been an x.

    Let's simplify the whole thing and see if we can get a cleaner statement of the relationship between x and k:

    log(base2)x+log(base2)(x+5)=k

    log(base2) [x(x+5)] = k {by properties of logarithms}

    That's an expression for k in terms of x

    x(x+5) = 2^k {definition of logarithms}

    x^2 + 5x = 2^k

    x^2 + 5x + 25/4 = 2^k + 25/4 {completing the square}

    (x + 5/2)^2 = 2^k + 25/4

    x + 5/2 = √(2^k + 25/4)

    (because a negative value won't satisfy the logarithm)

    x = - 5/2 + √(2^k + 25/4)

    That's x in terms of k.

    When k=7, we have

    x - 5/2 + √(2^k + 25/4) = - 5/2 + √(128 + 25/4)

    = - 5/2 + √(537/4) = (√537 - 5) / 2 = about 9.09

    Using the equation

    x^2 + 5x = 2^k

    above, we see that with k=7,

    x^2 + 5x - 128 = 0

    and the solution above does, indeed, fit.

  • 1 decade ago

    wtf?

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