logarithm in terms of x and y

Question

I wanted to know if I computed the following problem correctly?

"Given $\log3=x$ and $\log4=y$, find a solution to $\log144$ in terms of $x$ and $y$"

This is what I did:

$$\log3 + \log4 = x+y$$

$$\log(3\times4) =x+y$$

$$\log12= x+y$$

$$\log144 = \log12^2 = 2 \log12 = 2(x+y)$$

Therefore $$\log 144 = 2(x+y)$$

Answer 1

Yes, that is right!

Another way is $\log(144) = \log(16)+\log(9)=2\log4+2\log3=2y+2x$.

Answer 2

Yes that correct. To check ww can proceed backwards

$$\log (144)=\log (12^2)=2\log 12=2\log (3\cdot 4)=2(\log 3+\log 4)=2(x+y)$$