Log Integer, Integer Log

Algebra Level 2

$\log_6 (3x+1) = \log_{36} (mx^2+36x+1)$

Let $S$ denote the set of all integers $m$ such that the equation above only has non-negative integer roots (in $x$ ). Find the sum, $\sum \limits_{y \in S} |y|$ .

The answer is 66.

2 solutions

Guilherme Dela Corte
Jul 25, 2020

We can rewrite the equation as $\log_6 (3x+1) = \frac{\log_6 (mx^2+36x+1)}{2} \leftrightarrow 9x^2+6x+1=mx^2+36x+1 \leftrightarrow (9-m)x^2 = 30x \rightarrow x = 0 \text{ or } \frac{30}{9-m}$

Because the positive divisors of $30$ are $1, 2, 3, 5, 6, 10, 15, 30$ , we have that $m = {-21, -6, -1, 3, 4, 6, 7, 8}$ .

So our desired answer is $21 + 6 + 1 + 3 + 4 + 6 + 7 + 8 = \boxed{56.}$

I think the fact that it asks for the sum of all possible values of m's absolute value was a little confusing since 6 is counted twice. Although it helps that 50 isn't one of the options.

Tristan Goodman - 10 months, 3 weeks ago

@Guilherme Dela Corte @Chew-Seong Cheong I think the only accepted values for $m$ are $-6$ and $-21$ . Due to restrictions ( $3x+1>0$ and $m{{x}^{2}}+36x+1>0$ ) all the others values for $m$ are rejected, because either the RHS of the equation has no meaning, or $0$ is is a root that is not a positive integer (in the case $m=-1$ ). So the answer is wrong and you must edit the problem.

Thanos Petropoulos - 10 months, 2 weeks ago

For $m = -1$ , the root is $x = 3$ . I think you might be mistaken.

Guilherme Dela Corte - 10 months, 2 weeks ago

In fact, there is a problem with all suggested values of $m$ .
For the value $m=-1$ , for instance, it is true that the equation has $x=3$ as a solution, but it is not the only one. The value $x=0$ is a solution as well and this is not a positive integer, thus the requirement that the equation has ONLY POSITIVE integer roots is not satisfied, so the value $m=-1$ , as well as all the others, should be excluded. After all, the value $x=0$ is a trivial root of the original equation, no matter what the value of $m$ is.
I guess things can be settled, if you rephrase the first sentence of the problem as follows:

Let $m$ be an integer such that the equation above has only non-negative integer roots.

Finally, there is no reason for the value $\left| m \right|=6$ to be counted twice. All possible values of $\left| m \right|$ are $1$ , $3$ , $4$ , $6$ , $7$ , $8$ and $21$ , with a sum of $50$ .

Thanos Petropoulos - 10 months, 2 weeks ago

Chew-Seong Cheong
Jul 26, 2020

$\begin{aligned} \log_6 (3x+1) & = \log_{36} (mx^2+36x + 1) \\ \frac {\log (3x+1)}{\log 6} & = \frac {\log (mx^2+36x+1)}{2\log 6} \\ 2 \log (3x+1) & = \log (mx^2+36x+1) \\ \implies (3x+1)^2 & = mx^2+36x+1 \\ 9x^2 + 6x + 1 & = mx^2+36x+1 \\ (9-m)x^2 & = 30x & \small \blue{\text{For }x \ne 0} \\ (9-m) x & = 30 \\ \implies x & = \frac {30}{9-m} \end{aligned}$

For $x$ to be a positive integer, $9-m$ must be one of the eight positive divisors of $30 = 2 \cdot 3 \cdot 5$ , which are $1, 2, 3, 5, 6, 10, 15, 30$ , when $m = 8, 7, 6, 4, 3, - 1, -6, - 21$ and the sum of all $|m|$ is $8+7+6+4+3+1+6+21 = \boxed {56}$ .

Log Integer, Integer Log

The answer is 66.

2 solutions

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