Solving $y=ax^{b}$ with logarithms

Question

[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?

Answer 1

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.

Answer 2

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.

$$b=\frac{Y_1-Y_0}{X_1-X_0}.$$

Answer 3

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

$$a = \exp\left( \frac{\log_27 - \log_38 }{\log_3e - \log_2e} \right)$$