Our teacher said the answer is D but I don't get why A can't be the answer.

[kindred axle]

If someone can help me to understand that, I will really appreciate it

[loud plover]

You can only take the logarithm of a positive, so $x - 3 > 0$

So, option B is out.

The base must also be positive and can't be 1, so $x - \frac{5}{2} > 0$ and $x - \frac{5}{2} \neq 1$.

This restriction matches (A) and (C) but doesn't contradict (D) or (E).

Notice that (C) uses the a correct set notation for all values above 3 except for 7 / 2, while (A) does not.

We have yet to deal with the value of the logarithm being under 2.

If $x\in\left(3,,\frac{7}{2}\right)$, then we get expressions like:

Put $x = 3.01: \quad \log_{0.51}{0.01}=\frac{\log_{10}{0.01}}{\log_{10}{0.51}} = \frac{-2}{-0.29242...}=6.839...$

Put $x = 3.49: \quad \log_{0.991}{0.49}=\frac{\log_{10}{0.49}}{\log_{10}{0.991}} = \frac{-0.30980...}{-0.0043648...}=70.977...$

Clearly the domain $x\in\left(3,,\frac{7}{2}\right)$ is excluded, so we are down to (D) or (E).

Put $x = 3.51: \quad \log_{1.01}{0.51}=\frac{\log_{10}{0.51}}{\log_{10}{1.011}} = \frac{-0.2924298...}{0.00432137...}=-67.670...$

We have enough to conclude that (D) is the correct answer.

[young spire]

Jay

[young spire]

Jay

[loud plover]

Taking a more rigorous approach...

The expression inside the log must be positive $\implies x\in\left(3,,\infty\right)$

The base of the log must be positive $\implies x\in\left(\frac{5}{2},,\infty\right)$

But, the base of the log cannot be 1 $\implies x \neq \frac{7}{2}$

Combining these restrictions, the domain is:

$x\in\left(3,,\infty\right)\setminus\left{\frac{7}{2}\right}$ or $x\in\left(3,,\frac{7}{2}\right)\cup\left{\frac{7}{2},,\infty\right}$

Now, solving the inequality requires cases.

Case 1 - Base < 1 and so $3 < x < \frac{7}{2}$:

$\log_{\left(x - \frac{5}{2}\right)}{(x - 3)} < 2$

$\frac{\log_{10}{(x-3)}}{\log_{10}{\left(x - \frac{5}{2}\right)}} < 2$

$\log_{10}{(x-3)} > 2\log_{10}{\left(x - \frac{5}{2}\right)} \qquad \text{as the denominator was negative}$

$\log_{10}{(x-3)} > \log_{10}{\left(x - \frac{5}{2}\right)^2}$

$x - 3 > \left(x - \frac{5}{2}\right)^2$

$4x - 12 > 4x^2 - 20x + 25$

This rearranges to $4(x - 3)^2 + 1 < 0$, which is never true.

So, all values in the domain of Case 2 are invalid.

Case 2 - Base > 1 and so $x > \frac{7}{2}$:

$\log_{\left(x - \frac{5}{2}\right)}{(x - 3)} < 2 \implies x - 3 < \left(x - \frac{5}{2}\right)^2$

This rearranges to $0 < 4(x - 3)^2 + 1$ and is true for all real $x$.

So, all values in the domain of Case 2 are valid.

Taking the cases into account, the solution set is $x\in\left(\frac{7}{2},, \infty\right)$.

[young spire]

Jay