Daily math problems for October 20, 2026 — Algebra, Calculus, Fundamental Algebra & more

Use the product rule for logarithms: $${\log }_3(9x)={\log }_{3}9+{\log }_{3}x$$ Since $9=3^2$, we have ${\log }_{3}9=2$, and we are given ${\log }_{3}x=4$: $${\log }_3(9x)=2+4=6$$

$f$ is a polynomial (continuous on $[-2,2]$, differentiable on $(-2,2)$), and $f(-2)=0=f(2)$, so Rolle's hypotheses hold.

$f^{\mathrm{\prime }}(x)=2x$, so $f^{\mathrm{\prime }}(c)=0⟹c=0$.

${\displaystyle \frac{2x}{6}}={\displaystyle \frac{x}{3}}$.

$x^2-9=x^2-3^2=(x-3)(x+3)$ — the difference-of-squares identity.

$$\mid {\displaystyle \frac{a_{n+1}}{a_n}}\mid ={\displaystyle \frac{1\mathrm{/}(n+1)!}{1\mathrm{/}n!}}={\displaystyle \frac{n!}{(n+1)!}}={\displaystyle \frac{1}{n+1}}\mathrm{.}$$

Then $L={\displaystyle \underset{n\to \mathrm{\infty }}{\lim }{\displaystyle \frac{1}{n+1}}=0<1}$, so the series converges absolutely.

$V=\pi {\int }_0^4(\sqrt{x}{)}^{2}dx=\pi {\int }_0^{4}xdx=\pi \cdot {\displaystyle \frac{16}{2}}=8\pi$.

$\bullet$ Compute the modulus of the base:

$$\mathrm{\mid }1+i\sqrt{3}\mathrm{\mid }=\sqrt{1+3}=2$$

$\bullet$ Modulus distributes over powers:

$$\mathrm{\mid }(1+i\sqrt{3}{)}^{2024}\mathrm{\mid }=2^{2024}$$

$\bullet$ The denominator is real and positive, so:

$$\mid {\displaystyle \frac{(1+i\sqrt{3}{)}^{2024}}{2^{2023}}}\mid ={\displaystyle \frac{2^{2024}}{2^{2023}}}=2$$

Middle coefficient: $\left(\genfrac{}{}{0px}{}{8}{4}\right)=70$.

In $V_4$, every non-identity element has order $2$. (This distinguishes it from ${\mathbb{Z}}_4$, which has an element of order $4$.)

$\sin x+\cos x=\sqrt{2}\text{\hspace{0.17em}⁣}(\sin x\cdot {\displaystyle \frac{\sqrt{2}}{2}}+\cos x\cdot {\displaystyle \frac{\sqrt{2}}{2}})=\sqrt{2}\sin \text{\hspace{0.17em}⁣}(x+{\displaystyle \frac{\pi }{4}})$.

Since $\sin$ has maximum $1$, the maximum of $f$ is $\sqrt{2}$, attained at $x={\displaystyle \frac{\pi }{4}}$.