# Thread: Difficulty finding a, b,c

1. ## 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

2. 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.

3. Originally Posted by Moo
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.

4. 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

5. Originally Posted by chaneliman
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).

6. q1 i got up to this point, (2/ log2) log 2 +c = 10, what do i do now?

7. Originally Posted by chaneliman
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).

Originally Posted by mr fantastic
[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.