Determining logarithmic function of the form f(x) = k ln(ax+b)
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HallsofIvy
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In order to determine two values, a and b, you need two equations. You have two points marked, (0, 0) and (-1/2, -2). (You give only "x= -1/2" but it looks like y= -2 there.) Unfortunately there is a third undetermined constant, k.
The equation is given as y= k ln(ax+ b). With x= 0 and y= 0 that gives 0= kln(a(0)+ b)= k ln(b). With x= -1/2 and y= =2, -2= k ln(-a/2+b).
If k ln(b)= 0 either k= 0 or ln(b)= 0. If k= 0 then k ln(-a/2+ b)=0, not -2 we can't k=0. We must have ln(b)= 0 so b= 1.
Then ln(-a/2+ 1)= -2/k so -a/2+ 1= e^{-2/k}, -a/2= e^{-2/k}+ 1, and then a= -2e^{-2/k}-2.
That's the best we can do without either a third point or a specific value for k:
y= k ln((-2e^{-2/k}- 2)x+ 1).
The equation is given as y= k ln(ax+ b). With x= 0 and y= 0 that gives 0= kln(a(0)+ b)= k ln(b). With x= -1/2 and y= =2, -2= k ln(-a/2+b).
If k ln(b)= 0 either k= 0 or ln(b)= 0. If k= 0 then k ln(-a/2+ b)=0, not -2 we can't k=0. We must have ln(b)= 0 so b= 1.
Then ln(-a/2+ 1)= -2/k so -a/2+ 1= e^{-2/k}, -a/2= e^{-2/k}+ 1, and then a= -2e^{-2/k}-2.
That's the best we can do without either a third point or a specific value for k:
y= k ln((-2e^{-2/k}- 2)x+ 1).
Dr.Peterson
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As I read it, the asymptote is x = -1/2. You only need to find a and b, which relate only to the horizontal aspects of the graph; k relates to vertical aspects, of which we have no knowledge. So the question is solvable.
Consider the argument ax+b. What does the point (0, 0) tell you about this? What about the asymptote? That will give you two equations in a and b, which you can solve.
chunkytown
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and what are the equations ?
D
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Did you read response #3 - carefully? DrP. has described in detail a method to find those equations.
Please show us what you have tried and exactly where you are stuck.
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