How to solve logarithmic inequality?

Question

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.

Please help me to solve it.

Update

The base of the logoritm is 3.

Answer 1

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.

Answer 2

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$.

Answer 3

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 $\log_3(3-2x)$).

Answer 4

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:

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