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Thread: polynomials,indices,logarithms

  1. #1
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    polynomials,indices,logarithms

    1.
    Given the identity
    $\displaystyle x^4 + x + 1$ = ($\displaystyle x^2+ A$)($\displaystyle x^2-1$) + Bx + C

    determine the numerical value of A ,B and C. By giving x a suitable value, find the remainder when 100000101 is divided by 101.

    I know how to do the first part, A is 1, B is 1 and C is 2, but don't know how to do part 2 which is in blue colour.

    2.
    Solve $\displaystyle 2^{-x}$$\displaystyle -2(2)^{-(x/2)+1}$ +3 =0

    3.
    Solve the equation
    $\displaystyle log_{3}$(2x-3) = 2 - $\displaystyle log_{9}$$\displaystyle (2x-1)^2$
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  2. #2
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    Quote Originally Posted by wintersoltice View Post
    1.
    Given the identity
    $\displaystyle x^4 + x + 1$ = ($\displaystyle x^2+ A$)($\displaystyle x^2-1$) + Bx + C

    determine the numerical value of A ,B and C. By giving x a suitable value, find the remainder when 100000101 is divided by 101.

    I know how to do the first part, A is 1, B is 1 and C is 2, but don't know how to do part 2 which is in blue colour.
    It should be fairly obvious that the value of x you choose is x = 100 , and your diving by x+1, your values of A B and C are correct.leave the numbers out for now as it makes things simpler.

    start with $\displaystyle \frac{x^4 + x + 1}{x+1} = \frac{(x^2 + 1)(x^2 -1) + x -2 }{x+1} $

    $\displaystyle \frac{(x^2 + 1)(x+1)(x-1) + x -2 }{x+1} = (x^2 + 1)(x-1) + \frac{x-2}{x+1} $

    Can you take it form here ?

    2.
    Solve $\displaystyle 2^{-x}$$\displaystyle -2(2)^{-(x/2)+1}$ +3 =0
    write this as $\displaystyle 2^{-x}-4(2)^{-\frac{x}{2}} +3 =0$
    then let $\displaystyle (2)^{-\frac{x}{2}} =u$
    so the equation is $\displaystyle u^2 -4u +3 =0$


    3.
    Solve the equation
    $\displaystyle log_{3}$(2x-3) = 2 - $\displaystyle log_{9}$$\displaystyle (2x-1)^2$
    use the change of base on the log base 9 and consider writing 2 as $\displaystyle log_{3}(9) $


    Bobak
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