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?
$\displaystyle 2=a\log_2(5-b)$
$\displaystyle 4=a\log_2(7-b)$
The simplest way to eliminate a from those equations is to divide one by the other:
$\displaystyle \frac{4}{2}= 2= \frac{log_2(7- b)}{log_2(5- b)}$
or simply
$\displaystyle 2\log_2(5- b)= log_2(7- b)$
$\displaystyle log_2((5-b)^2= log_2(7- b)$
and, since logarithm is "one-to-one", $\displaystyle (5- b)^2= 7- b$, 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 for $\displaystyle a$ and $\displaystyle b$?
. . $\displaystyle \begin{array}{cccc}2&=& a\log_2(5-b) & [1] \\ \\[-3mm] 4 &=& a\log_2(7-b) & [2] \end{array}$
$\displaystyle \begin{array}{cccccc}
\text{Multiply [1] by 2:} & 2a\log_2(5-b) &=& 4 \\
\text{Equate to [2]:} & a\log_2(7-b) &=& 4 \end{array}$
We have: .$\displaystyle 2{\color{red}\rlap{/}}a\log_2(5-b) \;=\;{\color{red}\rlap{/}}a\log_2(7-b) \quad\Rightarrow\quad \log_2(5-b)^2 \:=\:\log_2(7-b)$
Then: .$\displaystyle (5-b)^2 \:=\:7-b \quad\Rightarrow\quad b^2 -9b + 18\:=\:0$
. . . . $\displaystyle (b-3)(b-6) \:=\:0 \quad\Rightarrow\quad b \:=\:3,\,6$
But $\displaystyle b=6$ is extraneous.
The only solution is: .$\displaystyle \boxed{b = 3}$
Substitute into [1]:
. . $\displaystyle a\log_2(5-3)\:=\:2 \quad\Rightarrow\quad a\log_2(2) \:=\:2 \quad\Rightarrow\quad a\cdot 1 \:=\:2 \quad\Rightarrow\quad\boxed{ a \:=\:2}$
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You are given the function: .$\displaystyle f(x) \;=\;\log_2(x-b)$
. . and two points: .$\displaystyle (5,2),\;(7,4)$
Right?