Doubt: AL-QLE-017-20260607-0223-q4

This doubt is really about the three conditions that make a logarithm defined at all — once you internalise those, every 'range of base' question becomes mechanical.

By definition, $\log_b a = y$ means $b^y = a$. For this to work as a proper function, three conditions must hold together: argument $a > 0$, base $b > 0$, and base $b \neq 1$.

Why each one? If $b = 1$, then $1^y = 1$ for every $y$, so $\log_1 a$ has no solution when $a \neq 1$ and infinitely many when $a = 1$ — useless either way. If $b = 0$ or $b < 0$, real powers aren't reliably defined: $0^{-1}$ blows up, and something like $(-2)^{1/2}$ isn't real. So the allowed range for a base is

$b \in (0,1) \cup (1,\infty).$

Note that $0 < b < 1$ is perfectly legal — e.g. $\log_{1/2} 8 = -3$ works fine; a fractional base just flips the inequality when you remove the log.

Now the 'why not 2 and 5' part. The base expression here has roots 2 and 5 — that's where those numbers come from: base $= (x-2)(x-5) = x^2 - 7x + 10$. Apply the base conditions to it.

Base $> 0$: $(x-2)(x-5) > 0$. A product of two factors is positive when both have the same sign, which gives $x < 2$ or $x > 5$. At exactly $x = 2$ and $x = 5$ the base equals $0$, and $\log_0(\cdot)$ does not exist — that is precisely why these two values are thrown out. For $2 < x < 5$, one factor is negative and the other positive, so the base is negative — rejected as well.

Base $\neq 1$: set $(x-2)(x-5) = 1 \Rightarrow x^2 - 7x + 9 = 0 \Rightarrow x = \frac{7 \pm \sqrt{13}}{2}$. These two values must also be dropped. They're irrational, so in integer-value versions of this question they don't remove any integer.

So the checklist for any 'is this log defined' question is always: argument $> 0$, base $> 0$, base $\neq 1$ — test every candidate $x$ against all three. The base-equals-1 check is the one almost everybody forgets.

Final answer: the base must lie in $(0,1) \cup (1,\infty)$; with base $(x-2)(x-5)$, valid $x$ are $x < 2$ or $x > 5$ (excluding $x = \frac{7 \pm \sqrt{13}}{2}$), and $x = 2, 5$ are rejected because they make the base $0$.