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Find a/b when 2 log(a - 3b) = log a + log b.

 Jul 1, 2022
 #1
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Find a/b if $\displaystyle 2\log{(a - 3b)} = \log a + \log b$

 Jul 1, 2022
 #2
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a/b = ((10^(log(ab)))/2b) + 3

Guest Jul 1, 2022
 #3
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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!

 Jul 3, 2022

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