Using logs properties: \(alog(x)=log(x^a)\) and \(log(a)+log(b)=log(ab)\) yields:
\(log(a-3b)^2=log(ab) \iff (a-3b)^2=ab\)
Expanding the left-hand side:
\(a^2-6ab+9b^2=ab\)
Simplify:
\(a^2-7ab+9b^2=0\)
Now, we want \(\dfrac{a}{b}\), so we divide this equation by \(b^2\) to get:
\((\dfrac{a}{b})^2-7(\dfrac{a}{b})+9=0\)
This is a quadratic in \(\dfrac{a}{b}\), which we can solve directly or by substitution as follows:
Let \(y=\dfrac{a}{b}\)
So our quadratic equation becomes: \(y^2-7y+9=0\)
Using the quadratic formula: \(y=\frac{7\pm \sqrt{49-36}}{2}=\frac{7\pm \sqrt{13}}{2}\)
But recall: \(a>0 , b>0\), and \(a-3b>0 \implies a>3b \implies \frac{a}{b} > 3\)
Thus, we reject: \(y_2=\frac{7-\sqrt{13}}{2}\) as it is less than 3.
Therefore, \(y_1=\dfrac{a}{b}=\frac{7+\sqrt{13}}{2}\) is the only solution.
Hope this helps!
