[y = 7 when x = 2 and y = 8 when x = 3] [a,b > 0 and are fixed real numbers]
I understand that this equation can be linearized using logarithms:
$$y = ax^b$$$$log(y) = log(ax^b)$$$$log(y) = log(a) + log(x^b)$$$$log(y) = b.log(x) + log(a)$$
And furthermore, given two equations rearranged for b:
$$b = log2(7) - log2(a)$$$$b = log3(8) - log3(a)$$
However, I'm still left with two unknowns and I haven't the foggiest how it is possible to solve for both of them.
I suppose my question in its most general form would be: How do I solve the rest of the unknowns for the equation y = ax^b when given two values of y and x?
$\endgroup$3 Answers
$\begingroup$We have the system of equations \begin{align} 7&=a(2^b)\tag1\\ 8&=a(3^b)\tag2\\ (2)\div(1): \frac87&=\left(\frac32\right)^b\\ b&=\log_{3/2}\left(\frac87\right)\\ \text{or}:\quad b&=\frac{\log8-\log7}{\log3-\log2} \end{align}
Once you have this, substitute into either $(1)$ or $(2)$ to get your answer.
$\endgroup$ 2 $\begingroup$If you write the equation in the form
$$\log y=\log a+b\log x,$$
this is linear in $\log x,\log y$, which you can write
$$A+bX=Y.$$
If you know two $X$ and two $Y$, you form the system
$$\begin{cases}A+bX_0=Y_0,\\A+bX_1=Y_1,\end{cases}$$
which you solve for instance by Cramer.
$\endgroup$ $\begingroup$$$b=\frac{Y_1-Y_0}{X_1-X_0}.$$
Continue with
$$b = \log_27 - \log_2a\tag 1$$$$b = \log_38 - \log_3a\tag 2$$
(1) = (2)
$$\log_38 - \log_ 3a = \log_27 - \log_2a$$
rearrange
$$ \log_3a -\log_2a= \log_27 - \log_38 $$
$$\ln a\left(\ln_3e - \ln_2e\right)=\log_27 - \log_38 $$
which yields
$$a = \exp\left( \frac{\log_27 - \log_38 }{\log_3e - \log_2e} \right)$$
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