Equal Logs

Algebra Level 4

$\log_9 p = \log_{12} q = \log_{16} (p + q)$

Suppose that $p$ and $q$ are positive numbers for which satisfy the equation above.

Then $q/p$ can be expressed in the form $\frac { x + \sqrt{y}}{z}$ , where $x, y, z$ are positive integers, and $y$ is not divisible by the square of a prime. Find $x + y + z$ .

The answer is 8.

1 solution

Alex Wang
Apr 15, 2015

Let $a = \log_9 p = \log_{12} q = \log_{16} (p + q).$

What we want is $q/p = (4/3)^a$ .

Then, $9^a + 12^a = 16^a$ .

Dividing both sides by $16^a$ , $(\frac{3}{4})^{2a} + (\frac{3}{4})^{a} -1=0.$

Using the quadratic formula, we get $(\frac{3}{4})^{a}= \frac{-1 + \sqrt{5}}{2}.$

What we want is the reciprocal, or $\frac{1 + \sqrt{5}}{2}.$

Our final answer is $1 + 5 + 2 = \boxed{8} .$

can we take log {16}(p+q) =a since we have already taken a as log {9}p ??

Nithin Nithu - 6 years, 1 month ago

They're equal?

Alex Wang - 6 years, 1 month ago

Oh I see what you mean now. Yes we can because the problem initially stated that they're equal.

Alex Wang - 5 years, 10 months ago

Equal Logs

The answer is 8.

1 solution

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