Proving that not defined value is equal to something

Question

My younger brother (9th Grader) got the following maths problem-

Given: $$2^a = 3^b = 6^c$$ Prove:

$$c=\frac{a * b}{a+b}$$

From my elementary knowledge of mathematics it seems like a=b=c=0.Also, (ab)/(a+b) is not defined and not defined can be equal to 0. Which makes me think if the question makes any sense. They could have also asked if (ab)/(a+b) = 182 i.e. some random number.

My question is if the output of

Not Defined == A Number

is true or false?

Does this question really makes sense?

Unfortunately the teacher is pretty arrogant and doesn't want to give an answer to this question!

Answer 1

Suppose that $2^a=3^b=6^c$, where $a\ne 0\ne b$. Take logs base $2$:

$$a=b\lg 3=c\lg 6=c(1+\lg 3)\;.$$

Then

$$\frac{ab}{a+b}=\frac{b^2\lg 3}{b+b\lg 3}=\frac{b\lg 3}{1+\lg 3}=\frac{a}{a/c}=c\;.$$

Of course the only solution with integral $a,b$, and $c$ is $a=b=c=0$, but there are certainly non-integral solutions.

Answer 2

$$2^a=3^b=6^c=k\text{(say)}$$

So, $2=k^{\frac1a},3=k^{\frac1b},6=k^{\frac1c}$

$$\implies k^{\frac1a} \cdot k^{\frac1b}=k^{\frac1c} $$

$$\implies k^{\frac1a+\frac1b} =k^{\frac1c} $$

We know, $a^m=a^n\implies m=n$ if $a\ne0,\pm1$

Here if $k=1, a=b=c=0$ but $\frac{ab}{a+b}=\frac00$ i.e., undefined