Find an equation of the form y=c+blogax whose graph contains the points (4,10) and (2,12)

Question

I'm completely stuck on this question. It's on our exam review worksheet and all they give us is the correct answer. They don't show how to solve this. I am able to do the y=ca^x problems just fine but this whole 3 variable one is really throwing me off. Does anyone know how to solve this?

Answer 1

There is more than one set of (a, b, c) which will work.

To start set up the equations

10 = c + b log( 4 a) and

12 = c + b log( 2 a)

Subtract the first one from the second one and use some simple properties of the log function to get

b = 2/ log(1/2) = -6.6438

Then, substituting for b, work with either the first one or the second one (they are equivalent) to get

10 = c - 6.6438 log (4a)

It easy to show than some solutions to this equation are

a = 1/4 c = 10

a = 1/2 c = 12

a = .707 c = 13

and so on.

Answer 2

I'm assuming the equation is: y = c + blog_a(x)

There are 3 variables to find (a, b, and c) but only 2 data points. That means you don't have enough data points to develop a solution using a system of equations. (3 unknowns requires 3 equations which would require 3 data points). So, let's select a = 2

I prefer to use the exponential form because I can typically solve them without a calculator so I'm going to convert the equation to an exponential:

y = c + blog_a(x) and since I selected a = 2:

y = c + blog₂(x)

y - c = blog₂(x)

(y - c)/b = log₂(x)

2^{(y - c)/b} = x

Plugging in the data points for (x, y) we get two equations in two unknowns:

(1) 2^{(10 - c)/b} = 4 or 2^{(10 - c)/b} = 2² meaning that (10 - c)/b = 2 or 10 - c = 2b

(2) 2^{(12 - c)/b} = 2 or 2^{(12 - c)/b} = 2¹ meaning 12 - c = b

Plugging "12 - c" into equation 1 in place of "b", we get:

10 - c = 2(12 - c)

10 - c = 24 - 2c

c = 14

Plugging that into equation 2, we get b = -2

So an equation that works is:

y = 14 - 2log₂(x)

There are others though that will work. Just select a different value for "a"