How do I prove that ab+5(a-b)=1

Question

If $\log_{12}18=a$ and $\log_{24}54=b$ then how do I prove $ab+ 5(a-b)=1$?

I figured that out it's $\log_ab$ and $\log_{2a}3b$ but how do I solve it?

Answer 1

$12^a = 18, 24^b = 54 \to 24^b\cdot 12^{-a}= 3 \to 2^{3b}\cdot 3^b\cdot 2^{-2a}\cdot 3^{-a} = 3 \to 2^{3b-2a} = 3^{1+a-b} \to (3b-2a)\ln 2 = (1+a-b)\ln 3$. This is what I preliminarily got. More on this later for you.What you need to prove is: $a = \dfrac{5b+1}{b+5}$.Can you use my hint to connect with the answer?