logs

Find a/b if $\displaystyle 2\log{(a - 3b)} = \log a + \log b$

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!