# How to solve logarithmic inequality?

I have this function:

I need to configure the definition of the function domain:

But I stucked and I dont know how to solve inequality above.

Update

The base of the logoritm is 3.

• Make cases 1)where both are negative 2)both positive and solve – Archis Welankar Oct 8 '16 at 14:49
• Perhaps the question should be edited to clarify: $f(x) = \sqrt{(1-x)\log_3(3-2x)}.$ – B. Goddard Oct 8 '16 at 15:48

There are two things to consider. In the real numbers, we don't take square roots of negatives and we don't take $\log$'s of negatives or zero. So first we need $3-2x >0$ so we have $$x<3/2.$$

Next we need the product $\log(3-2x)(1-x) \geq 0$. Both factors must have the same sign. If they are both non-negative , then $\log(3-2x)\geq 0$ and $1-x\geq 0$, so that $3-2x\geq 1$ and $x\leq 1$. Both of these reduce to $$x\leq 1.$$

If both are non-positive, then we reverse the inequalities in the paragraph above to get $$x\geq 1.$$

Thus the square root consideration really tells us nothing. The $\log$ consideration tell us $x<3/2$, and that's the domain.

• if you let x = 1.4, you obtain an illegal value. – Joseph Wood Oct 8 '16 at 15:29
• @JosephWood I don't think so. $\log(0.2)$ is negative and $1-1.4$ is negative, so their product is positive and the square root is defined. – B. Goddard Oct 8 '16 at 15:46
• You are not comparing the same things. Yes it is true that when isolated $\log(0.2)$ is negative and equally true that $(1-1.4)$ is negative. However, we don't have $(\log(0.2)) * (-0.4)$, we have $\log(0.2)(-0.4) = \log(-0.08)$. See my solution below. – Joseph Wood Oct 8 '16 at 15:49
• @JosephWood The OP has stated that the factor $1-x$ is NOT inside the log. At face value, convention also would say that the factor is NOT inside the log. – B. Goddard Oct 8 '16 at 15:50
• Thanks for shedding some light on this topic. Had it not been for your astute attention, I would have simply answered this question and walked away learning nothing. Also, very nice explanation! – Joseph Wood Oct 9 '16 at 12:05

Hint:

$\log x <0$ when $x\in (0,1)$ and $\log x\geq 0$ when $x\geq 1$

This means that you have to solve quadratic inequality $(3-2x)(1-x)\geq 1$.

• How did you get this quadratic inequality (3−2x)(1−x)≥1? – Michael Oct 8 '16 at 14:54
• Argument of logarithmic function must be $\geq 1$. Argument in your case is $(3-2x)(1-x)$. – MaliMish Oct 8 '16 at 14:56
• but as I understand only (3-2x) is logoritmic funticon the (1−x) is linear function.Isn't it? – Michael Oct 8 '16 at 15:11
• @Joseph I interpreted the question as if $y:=(3-2x)(1-x)$ was argument. In that case we want $\log y\geq 1$. – MaliMish Oct 8 '16 at 15:48
• @MaliMish, my mistake... I misinterpreted what you wrote. Nice job!! +1 – Joseph Wood Oct 8 '16 at 15:58

This depends on how you interpret $$\log_3(3-2x)(1-x)$$ which is indeed ambiguous.

If you interpret it as $$\log_3\bigl((3-2x)(1-x)\bigr)$$ then the condition is $$(3-2x)(1-x)\ge1$$ so the logarithm is nonnegative.

If you interpret it as $$(1-x)\log_3(3-2x)$$ then the condition is split into two: $$\begin{cases} 1-x\ge0 \\[4px] 3-2x\ge1 \end{cases} \qquad\text{or}\qquad \begin{cases} 1-x<0 \\[4px] 0<3-2x<1 \end{cases}$$ (the former for both factors nonnegative, the latter for both factors negative, plus the existence of $\log_3(3-2x)$).

• Thanks for putting both interpretations. This seems to be quite a point of contention. – Joseph Wood Oct 8 '16 at 15:53

Using the properties of logarithm, we can split the initial term into two parts (Thanks to @egreg for pointing out the point about absolute values that is often overlooked but very important):

$$\log(3 - 2x)(1 - x) \geq 0 \implies \log|3 - 2x| + \log|1 - x| \geq 0 \implies$$

$$\log|3 - 2x| \geq -\log|1 - x| \implies \log|3 - 2x| \geq \log|1 - x|^{-1} \implies$$

$$3^{\log|3 - 2x|} \geq 3^{\log|1 - x|^{-1}} \implies |3 - 2x| \geq |1 - x|^{-1} \implies$$

$$|3-2x||1-x| \geq 1$$

Now we have to do a little work to determine the correct domain. Below we set up our number line with the "turning points" (i.e. the values that make each expression equal to zero) to establish our intervals (The negatives to the left of $1$ mean both expressions are negative, the $+-$ in the middle mean that the $(1-x)$ term is positive and the other expression is negative, and anything to the right of $\frac{3}{2}$, means both expressions are positive.

$$<-----------\space\space 1 \space ++++---- \space\space \frac{3}{2} \space ++++++++++>$$

For values of $x < 1$ we have:

$$\bigl(-(3-2x)\bigr)\bigl(-(1-x)\bigr) \geq 1 \implies (3-2x)(1-x) \geq 1$$ $$(2x - 1)(x - 2) \geq 0 \implies x \geq 2 \space\space\space x \leq 1/2$$

From the restriction above, we must have $x \leq 1/2$.

For the middle interval (i.e. $1 < x < 3/2$), we have:

$$\bigl(-(3-2x)\bigr)(1-x) \geq 1 \implies (3-2x)(1-x) \leq -1 \implies 2x^{2}-5x+4 \leq 0 \implies$$

$$x = \frac{5 \pm \sqrt{25 - 32}}{4} \implies x \in \mathbb{C}$$

Thus, there is no real solution on this interval.

And finally for our last interval, we have exactly the same thing as the first interval above, but with the caveat that $x > 3/2$. Thus we obtain $x \geq 2$.

So, the domain is $(-\infty, \tfrac{1}{2} \bigr] \cup [2, \infty)$.

Update
I have searched high and low for a logarithm problem set-up like this one and cannot find any exact matches. However, after viewing hundreds of logarithm problems, I'm inclined to lean toward NOT including the $(1-x)$ in the argument of the logarithm. In which case, the solutions provided by @B,Goddard or @egreg will suffice. Typically in textbooks, this problem would have appeared like one of the following to eliminate any ambiguity:

$$\bigl[\log(3 - 2x)\bigr] (1 - x) \quad \text{equivalently} \quad (1-x)\log(3-2x)$$ $$(i.e. (1 - x) \space \text{is not part of the argument for the logarithm})$$ $$-\text{OR}-$$ $$\log\bigl((3 - 2x)(1 - x)\bigr) \quad (i.e. (1 - x) \space \text{is part of the argument for the logarithm})$$

• There is some confusion here about notation. As written, the factor $1-x$ is not part of the argument for $\log$. Also, in a comment below MaliMish's solution, the OP says specifically that the factor is not part of the argument for $\log.$ – B. Goddard Oct 8 '16 at 15:55
• @B.Goddard, I think even the OP is a little confused as well. He does finish his comment with the question ...Isn't it?. – Joseph Wood Oct 8 '16 at 15:57
• Actually, $\log\bigl((3-2x)(1-x)\bigr)=\log\lvert 3-2x\rvert+\log\lvert 1-x\rvert$ – egreg Oct 8 '16 at 16:18
• @egreg, thanks for pointing this out. I've updated my answer. – Joseph Wood Oct 8 '16 at 17:27