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# logs

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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