Daily math problems for September 22, 2026 — Logs, Analytic Geometry, Quadratic Function & more

$\bullet$ Domain: both arguments positive, so $x>2$.

$\bullet$ Combine logs:

$${\log }_3((x-2)(x+6))=2$$

$\bullet$ Exponentiate and expand:

$$(x-2)(x+6)=9$$

$$x^2+4x-21=0$$

$\bullet$ Factor:

$$(x+7)(x-3)=0$$

$\bullet$ $x=-7$ fails the domain $x>2$. Only $x=3$ remains.

$\bullet$ Compute coordinates of $A$. Note ${\log }_{4}8=\frac{3}{2}$ and treat ${\log }_{1337}\mathrm{\mid }i\mathrm{\mid }=0$:

$$A=(2\cdot \frac{3}{2},0)=(3,0)$$

$\bullet$ Coordinates of $B$ from ${\log }_{2}32=5$ and $P_2=2$:

$$B=(5,2)$$

$\bullet$ Coordinates of $C$ from $0!=1$ and ${\log }_{2}512=9$:

$$C=(1,9)$$

$\bullet$ Apply the area formula:

$$\mathcal{A}=\frac{1}{2}\mathrm{\mid }{x}_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\mathrm{\mid }$$

$$=\frac{1}{2}\mathrm{\mid }3(-7)+5(9)+1(-2)\mathrm{\mid }$$

$$=\frac{1}{2}\cdot 22=11$$

$(x-2)(x+2)<0\iff x\in (-2,2)$.

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{(1+x{)}^r-1}{x}}=r}$. (This generalizes the bonomial-derivative-at-zero.)

$\mathrm{\mid }{A}_n\mathrm{\mid }={\displaystyle \frac{n!}{2}}$ (one of the standard properties of the alternating group).

$0\le I_n\le {\displaystyle \frac{1}{n+1}}\to 0$, so $I_n\to 0$ by the squeeze theorem.

${\log }_{2}6+{\log }_2{\displaystyle \frac{8}{3}}={\log }_2\text{\hspace{0.17em}⁣}(6\cdot {\displaystyle \frac{8}{3}})={\log }_{2}16=4$.

Vertical: $x=\pm 1$ (two). Horizontal: $\lim f(x)=1$ (one). Total: $3$. (No slant — the rational function has equal-degree numerator and denominator.)

By Viète: $x_1+x_2=5$, $x_1{x}_2=4$. Hence $x_1^2+x_2^2=25-8=17$.

A ring homomorphism preserves both operations: addition AND multiplication.