A unique pair of integers a and b make the equation below true:12^a·18^b=6^(a+2)·27^b

This question is extremely tough, and I had to think about this one for a while before I was able to come up with any answer at all.

 

First, we need to simplify this equation because there is a lot of simplification that is possible. I will do this by converting as many bases as possible to common bases so that the equation is easier to digest.

 

\(12^a*18^b=6^{a+2}*27^b\\ (6*2)^a*(9*2)^b=6^a*6^2*(9*3)^b\\ 6^a*2^a*9^b*2^b=6^a*36*9^b*3^b\\ 2^a*2^b={\color{red}36}*3^b\\ 2^a*2^b={\color{red}2^2*3^2}*3^b\\ \frac{2^a*2^b}{2^2}=3^b*3^2\\ 2^{a+b-2}=3^{b+2} \)

 

I simplified this equation as much as possible to see if I could make a breakthrough. With nasty equations like this one, it often best to make a key observation. This key observation will lead you to find the answer quite straightforwardly. In order to make this key observation. To make this observation, I simplified this equation a little bit more. Let's consider \(2^x=3^y\). Is there a value of x and y that will make this equation hold true? If you think about it, then the answer might jump right out at you: x = 0 and y = 0. There are no other options because the bases are different. 

 

If \(2^{\color{red}0}=3^{\color{blue}0}\) and \(2^{\color{red}{a+b-2}}=3^{\color{blue}{b+2}}\), then by parallel structure of these equations, \({\color{red}a+b-2=0}\text{ and }{\color{blue}b+2=0}\). This is a simple system of equations, so solving this will not be too difficult.

 

\(b+2=0\\ b=-2\\ a+b-2=0\\ a-2-2=0\\ a-4=0\\ a=4\)

 

Therefore, the only unique pair of integers is \(a=4\text{ and }b=-2\)