Solve the equation: $12^{r-1}=7r$

Question

$12^{r-1}=7r$. I teach an Algebra 2 class, and I came across this question in one of their homework sets on logarithms. Surely this is a typo and should read $12^{r-1}=7$. I'm not sure how to solve it the other way around. Could anyone point me in the right direction on this problem?

Answer 1

Equations of the form $a^x-bx=0$ can be solved using the Lambert W function. Assuming a,b are positive, we have $$ e^{x\log(a)}=\frac{b}{\log(a)}x\log(a). $$ Putting $y=-x\log(a)$ we obtain $$ e^{-y}=-\frac{b}{\log(a)}y, $$ so that $$ ye^y=-\frac{\log(a)}{b}. $$ Then we have $y=W(-\frac{\log(a)}{b})$, which gives $$ x=-\frac{W(-\frac{\log(a)}{b})}{\log(a)}. $$ Above we could multiply the equation by $12$ to obtain $$ 12^x=72 x. $$

Answer 2

Seeing it is Algebra 2, another alternative to a typo is that the equation $$ 12^{r-1}=7r $$ is correct, but not over the real numbers, but for a finite field $\mathbb{F}_p$. Then we would read it as $$ 12^{r-1}\equiv 7r \bmod p. $$ For example, with $p=5$ we have a solution $r=4$, because $$ 1728=12^{3}\equiv 7\cdot 4 =28\bmod 5. $$