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Thread: help with a few problems (don't need to answer all)

  1. #1
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    help with a few problems (don't need to answer all)

    i have tried these many times and still can't get them.

    1) f(x)= ((5x^3)-(5x^2)+9x-9) / (x-1) when x < 1
    f(x)= (6x^2)-1x+a when x>=1

    What value is a to make the function continuous at 1?

    I tried making them equal to each other and putting x=1 but that didn't work.

    2) (e^2x) - (7e^x) + 6 = 0

    It says there are 2 solutions but I keep getting -33.1255 and that's not even one of them.

    3) y=loga(x) with the subscript being a. It passes through (28,1).

    What is the value of a?

    I have no idea what to do on this one.

    4) log2(x) + log2(x+3) = k

    Solve for x in terms of k.

    Again, I have no idea what to do. I can solve it if k is an integer but thats about it.

    Any help is greatly appreciated.
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  2. #2
    MHF Contributor
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    1. Your approach would normally work, except that the function $\displaystyle \frac{5x^3 - 5x^2 + 9x - 9}{x - 1}$ has a discontinuity at x = 1. But what is the limit of the function? Factor the numerator to yield $\displaystyle \frac{(5x^2 + 9)(x-1)}{x-1}$. Then cancel the factors of $\displaystyle x-1$ and you are left with $\displaystyle 5x^2 + 9$. At x = 1, this function is equal to 14, so the limit is 14, and that is the function value you want for the other part of your function.

    2. Substitute $\displaystyle u = e^x$, yielding $\displaystyle u^2 - 7u + 6 = 0$. Then solve for u by factoring and identifying the roots: $\displaystyle (u - 6)(u - 1) = 0$, so u = 1 or 6. Hence $\displaystyle e^x = 1$ or $\displaystyle e^x = 6$. From $\displaystyle e^x = 1$ you have the solution x = 0, and from $\displaystyle e^x = 6$ you have the solution x = ln 6.

    3. This is an easy one. The equation $\displaystyle y = \log_a x$ can be rewritten as $\displaystyle a^y = x$. With (28, 1) on the graph, we have $\displaystyle a^1 = 28$, or $\displaystyle a = 28$.

    4. You can combine the two logarithms to yield $\displaystyle \log_2 (x(x+3)) = k$, which can be written $\displaystyle 2^k = x(x + 3)$, or $\displaystyle x^2 + 3x - 2^k = 0$. Now apply the quadratic formula: $\displaystyle x = \frac{-3 \pm \sqrt{9 + 4\cdot 2^k}}{2}$. Recognizing that x must be a positive number in order to evaluate the logarithm, we have $\displaystyle x = \frac{-3 + \sqrt{9 + 2^{k+2}}}{2}$.
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  3. #3
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    ooooooo. Those explanations were great. Thanks a lot.
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