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Math Help - Difficulty finding a, b,c

  1. #1
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    Difficulty finding a, b,c

    1. The points (3,10) and (5,12) lie on the graph of the function with rule y=a loge (x-b) +c. The graph has a verticle asymptote with equation x=1. find the values of a,b,c

    2. The graph of the function with rule f(x)= a loge (-x)+b passes through the points (-2,6) and (-4,8). Find the values of a and b

    Thanks
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  2. #2
    Moo
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    Hello,

    f(x)=a \log_e (x-b)+c

    Verticle asymptote at x=1 means that \lim_{x \to 1}=+ \text{\ or \ } - \infty

    Assuming that x>b (or else, the function is undefined), the only way to have such a limit at x=1 is that 1-b=0

    -> b=1

    Then, replace the coordinates of the points to get a and c.
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  3. #3
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    Quote Originally Posted by Moo View Post
    Hello,

    f(x)=a \log_e (x-b)+c

    Verticle asymptote at x=1 means that \lim_{x \to 1}=+ \text{\ or \ } - \infty

    Assuming that x>b (or else, the function is undefined), the only way to have such a limit at x=1 is that 1-b=0

    -> b=1

    Then, replace the coordinates of the points to get a and c.
    In other words, solve the following two equations simultaneously for a and c:

    10 = a \log_e (2) + c .... (1)

    12 = a \log_e (4) + c \Rightarrow 12 = 2a \log_e (2) + c .... (2)

    I'd solve via 2 times (1) - (2) .....


    The same approach is used for the second question.
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  4. #4
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    i got for q1, 10/12 aloge (12-4)+c
    then 5/6= aloge (8)+c

    and for q2, i got a as 2/loge (2) but i can't sub that into the equation to get b=4
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  5. #5
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    Quote Originally Posted by chaneliman View Post
    i got for q1, 10/12 aloge (12-4)+c
    then 5/6= aloge (8)+c

    and for q2, i got a as 2/loge (2) but i can't sub that into the equation to get b=4
    Q1 Solve it in the way I advised. What you've posted is wrong.

    Q2 6 = a \log_e (2) + b .... (1)

    8 = a \log_e (4) + c \Rightarrow 8 = 2a \log_e (2) + c .... (2)

    Solve simultaneously using the elimination technique: 2 times (1) - (2).
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  6. #6
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    q1 i got up to this point, (2/ log2) log 2 +c = 10, what do i do now?
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  7. #7
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    Quote Originally Posted by chaneliman View Post
    q1 i got up to this point, (2/ log2) log 2 +c = 10, what do i do now?
    Wrong (and I don't know how you got it).

    Quote Originally Posted by mr fantastic View Post
    [snip]
    10 = a \log_e (2) + c .... (1)

    12 = a \log_e (4) + c \Rightarrow 12 = 2a \log_e (2) + c .... (2)

    I'd solve via 2 times (1) - (2)
    [snip]
    2 times (1) gives 20 = 2a \log_e (2) + 2c.

    Therefore 2 times (1) - (2) gives:

    20 - 12 = [2a \log_e (2) + 2c] - [2a \log_e (2) + c]

    \Rightarrow 8 = c.

    Now substitute c = 8 into either of (1) and (2) and solve for a.
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