Solving $79=A\log_{10}(2B+C)$, $72=A\log_{10}(4B+C)$, $62=A\log_{10}(6B+C)$

Question

For a question, I was given a set of $x$ and $y$ values to substitute into the equation $y=A\log_{10}(Bx+C)$ and solve for $A, B$, and $C$.

So I tried to type this into my calculator and solve for A, B, and C by using the x and y values I was given.

$$79=A\log(2B+C)\\72=A\log(4B+C)\\62=A\log(6B+C)$$

However, when I press execute to solve the equations in my calculator, it is continuously solving and never solves the problem.

Have I typed it incorrectly somewhere or something?

Answer 1

From the initial system

$$2B+C=10^{79/A}=\alpha^{79}\\4B+C=10^{72/A}=\alpha^{72}\\6B+C=10^{62/A}=\alpha^{62}$$ where $\alpha:=10^{1/A}$.

We eliminate $B,C$ with

$$\alpha^{79}-2\alpha^{72}+\alpha^{62}=0$$ or

$$\alpha^{17}-2\alpha^{10}+1=0.$$

This polynomial equation has three real roots, among which $1$ and $-0.91160237164296303133\cdots$ which we reject, and $1.0433225893499999029\cdots$ which gives

$$A=54.2929119083438\cdots$$

$B$ and $C$ easily follow.