Daily math problems for September 9, 2026 — Algebra, Vectors, Complex Numbers & more

$\bullet$ Substitute $t=3^x>0$, noting $9^x=t^2$ and $3^{x-1}=t\mathrm{/}3$:

$$t^2-\frac{10}{3}t+1=0$$

$\bullet$ Multiply by $3$:

$$3t^2-10t+3=0$$

$\bullet$ Factor:

$$(3t-1)(t-3)=0\Rightarrow t\in \{\frac{1}{3},3\}$$

$\bullet$ Back-substitute:

$$3^x=\frac{1}{3}\Rightarrow x=-1$$

$$3^x=3\Rightarrow x=1$$

$\bullet$ So $x\in \{-1,1\}$.

$\vec{u}\cdot \vec{v}=4m+3(m+1)=7m+3=0\Rightarrow m=-{\displaystyle \frac{3}{7}}$.

${\displaystyle \frac{1+i}{1-i}}\cdot {\displaystyle \frac{1+i}{1+i}}={\displaystyle \frac{(1+i{)}^2}{1-i^2}}={\displaystyle \frac{2i}{2}}=i$.

$2\sin x\cos x-\sin x=0\Rightarrow \sin x(2\cos x-1)=0$. $\sin x=0$: $x\in \{0,\pi \}$. $\cos x=1\mathrm{/}2$: $x\in \{\pi \mathrm{/}3,5\pi \mathrm{/}3\}$. Total: $4$.

$A=\left(\begin{array}{cc}2& 3\\ 1& -1\end{array}\right)$ — coefficients of $x$ and $y$ in each equation.

$\bullet$ Expand along the first row:

$$\det A=m(m^2-1)-1\cdot (m-0)+0$$

$\bullet$ Simplify:

$$\det A=m^3-2m=m(m^2-2)$$

$\bullet$ Set to zero:

$$m(m^2-2)=0$$

$\bullet$ Solutions:

$$m\in \{-\sqrt{2},0,\sqrt{2}\}$$

$(\sin x+\cos x{)}^2={\displaystyle \frac{1}{4}}$. Expanding: ${\sin }^{2}x+2\sin x\cos x+{\cos }^{2}x={\displaystyle \frac{1}{4}}$, so $1+2\sin x\cos x={\displaystyle \frac{1}{4}}$. Hence $\sin x\cos x=-{\displaystyle \frac{3}{8}}$.

By Stirling, $\sqrt[n]{n!}\sim {\displaystyle \frac{n}{e}}(2\pi n{)}^{1\mathrm{/}(2n)}$. The factor $(2\pi n{)}^{1\mathrm{/}(2n)}\to 1$, so ${\displaystyle \frac{\sqrt[n]{n!}}{n}}\to {\displaystyle \frac{1}{e}}$.

Using the closed forms $\cos {36}^{\circ }=\frac{1+\sqrt{5}}{4}$ and $\cos {72}^{\circ }=\frac{\sqrt{5}-1}{4}$, the difference is $\frac{(1+\sqrt{5})-(\sqrt{5}-1)}{4}=\frac{2}{4}=\frac{1}{2}$.

$S_n={\displaystyle \frac{n(3n+7)}{2}}$. Compute: $S_{17}={\displaystyle \frac{17\cdot 58}{2}}=493<500$ and $S_{18}={\displaystyle \frac{18\cdot 61}{2}}=549\ge 500$. Hence the smallest $n$ is $18$.