The simplest way to eliminate a from those equations is to divide one by the other:Originally Posted by user_5
I'm trying to determine the equation of a graph, and I have come up with these two equations, how do I solve for a and b?
}{log_2(5- b)})
or simply
and, since logarithm is "one-to-one",, a quadratic equation for b.
Hello, user_5!
Another approach . . .
I'm trying to determine the equation of a graph. .**
I have come up with these two equations.
How do I solve forand
?
. .![\begin{array}{cccc}2&=& a\log_2(5-b) & [1] \\ \\[-3mm] 4 &=& a\log_2(7-b) & [2] \end{array}](http://latex.codecogs.com/png.latex?\begin{array}{cccc}2&=& a\log_2(5-b) & [1] \\ \\[-3mm] 4 &=& a\log_2(7-b) & [2] \end{array})
We have: .
Then: .
. . . .
Butis extraneous.
The only solution is: .
Substitute into [1]:
. .
~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~
**
You are given the function: .
. . and two points: .
Right?
  |
LinkBack URL
About LinkBacks