# Solutions for this logarithmic equation.

For which values of $k$ does the equation

$\log_a(kx+3)+\log_a(x+1)=\log_a(2x+1)$

have one or more solutions in $x$?

The logarithmic functions must have the restriction that the argument is greater than zero (i.e. no complex numbers allowed).

That leaves these restrictions: $kx>-3$, $x>-1$, $x>-1/2$

What I have tried so far is as follows:

$\log_a((kx+3)(x+1))=\log_a(2x+1)$ - Logarithm laws

$(kx+3)(x+1)=2x+1$ - Logarithm laws

$kx^2+(k+3)x+3=2x+1$ - Expand

$kx^2+(k+1)x+2=0$ - Make a quadratic in $x$

Discriminant of this is a quadratic in $k$: $(k+1)^2-8k$

$k^2-6k+1\ge0$

So that means $k\le(3-2\sqrt2)$ or $k\ge(3+2\sqrt2)$

However, this doesn't take into account the original restrictions of the problem, so I've gotten stuck here. How might I improve this bound on $k$?

-
Why don't you just calculate $x$ as well and check the restrictions? –  Fabian Jan 12 '12 at 19:46
Alright thanks for the responses, I'll try that. –  user22837 Jan 12 '12 at 19:53

Your work so far is fine. However, these conditions, while necessary, need not be sufficient. For example, the conditions given allow $k=0$, but if $k=0$, then the equation reduces to $\log_a(3x+3) = \log_a(2x+1)$, hence $3x+3=2x+1$, hence $x=-2$, but if $x=-2$ then $\log_a(x+1)$ does not make sense any more.

Instead, let's focus on $kx+3\gt 0$ and $x\gt -\frac{1}{2}$ (note that the latter already implies $x\gt -1$, so we get that one "for free").

As you do, we conclude that we must have $$kx^2 +(k+1)x + 2 = 0,$$ hence $$x = \frac{-(k+1)+\sqrt{k^2-6k+1}}{2k}\quad\text{or}\quad \frac{-(k+1)-\sqrt{k^2-6k+1}}{2k}.$$

This leads to your necessary condition, $k\leq 3-2\sqrt{2}$ or $k\geq 3+2\sqrt{2}$. Now, what do we need for $x$ to be greater than $-\frac{1}{2}$?

If $k\gt 0$, then $(k+1)^2 \gt k^2-6k+1$, so $-(k+1)+\sqrt{k^2-6k+1}\lt 0$; both values of $x$ will be negative; in order for the smaller value to be greater than $-\frac{1}{2}$ we would need \begin{align*} \frac{-(k+1)-\sqrt{k^2-6k+1}}{2k}&\gt-\frac{1}{2},\\ -(k+1)-\sqrt{k^2-6k+1}&\gt -k &\text{(as }k\gt 0\text{)}\\ -1&\gt \sqrt{k^2-6k+1} \end{align*} which is impossible. Proceeding the same way with the other value of $x$ we get to \begin{align*} -1 & \gt -\sqrt{k^2-6k+1}\\ 1 &\lt \sqrt{k^2-6k+1}\\ 1 &\lt k^2-6k+1\\ 0 &\lt k(k-6) \end{align*} which mean that we must have $k\lt 0$ or $k\gt 6$; and since we are assuming $k\gt 0$, we conclude that we require $k\gt 6$. So if $k\gt 0$, then the only time we have $x\gt-\frac{1}{2}$ is when $$k\gt 6\quad\text{and}\quad x = \frac{-(k+1)+\sqrt{k^2-6k+1}}{2k}.$$ In addition, note that if $k\gt 6$, then the square root is defined.

We still need to check the final condition, $kx+3\gt 0$. We have \begin{align*} kx + 3 = \frac{-(k+1)+\sqrt{k^2-6k+1}}{2}+3 &\gt 0\\ -(k+1)+\sqrt{k^2-6k+1} + 6&\gt 0\\ -k + \sqrt{k^2-6k+1}+5&\gt 0\\ \sqrt{k^2-6k+1}&\gt k-5\\ k^2-6k+1 &\gt k^2-10k+25 &\text{(since }k-5\gt 0\text{)}\\ 4k &\gt 24\\ k&\gt 6 \end{align*} which is what we are already assuming. So $kx+3\gt 0$ will hold as well provided $k\gt 6$.

So, for positive $k$, the conditions for a solution to exist boil down to $k\gt 6$, in which case the only solution for $x$ is $$x = \frac{-(k+1)+\sqrt{k^2-6k+1}}{2k}.$$

Now, what about $k\lt 0$ (note all those values make $k^2-6k+1$ positive)?

If $k\lt 0$, then $(k+1)^2 \lt k^2-6k+1$, so $|k+1|\lt \sqrt{k^2-6k+1}$, hence $-\sqrt{k^2-6k+1}\lt k+1 \lt \sqrt{k^2-6k+1}$. Therefore, $$-(k+1)-\sqrt{k^2-6k+1} \lt 0 \lt -(k+1)+\sqrt{k^2-6k+1}.$$ Therefore, dividing by the negative number $2k$, we obtain that $$\frac{-(k+1)-\sqrt{k^2-6k+1}}{2k}\gt 0 \gt \frac{-(k+1)+\sqrt{k^2-6k+1}}{2k}.$$

So one value of $x$ is positive and the other is negative always. The negative value will always satisfy $kx+3\gt 0$, while the positive value will always satisfy $x\gt -\frac{1}{2}$.

Let's check when the negative value is greater than $-\frac{1}{2}$: \begin{align*} \frac{-(k+1)+\sqrt{k^2-6k+1}}{2k} &\gt -\frac{1}{2}\\ -(k+1)+\sqrt{k^2-6k+1} &\lt -k\\ \sqrt{k^2-6k+1} &\lt 1\\ k^2-6k &\lt 0\\ k(k-6) &\lt 0 \end{align*} which requires $0\lt k\lt 6$; but this is impossible when $k\lt 0$. So the negative value is never greater than $-\frac{1}{2}$, so the negative value of $x$ never yields a valid solution.

Let's now check when the positive value satisfies $kx+3\gt 0$: \begin{align*} k\left(\frac{-(k+1)-\sqrt{k^2-6k+1}}{2k}\right) + 3 &\gt 0\\ -(k+1)-\sqrt{k^2-6k+1} + 6 &\gt 0\\ -\sqrt{k^2-6k+1}&\gt k-5\\ k^2-6k+1 &\lt k^2 - 10k + 25\\ 4k &\lt 24\\ k &\lt 6 \end{align*} which always holds since $k\lt 0$. So the positive solution always satisfies the required conditions.

In summary, when $k\lt 0$, we always have a solution, and it will be $$x = \frac{-(k+1)-\sqrt{k^2-6k+1}}{2k}.$$

So it seems to me that the values of $k$ for which there is at least one real solution $x$ are $k\lt 0$ or $k\gt 6$.

-
+1 Great answer! –  user22837 Jan 13 '12 at 3:52