Daily math problems for August 10, 2026 — Logic, Calculus, Logs & more

$A\cap B$ contains the elements common to both sets: $\{2,3,5\}$.

For an odd function $f$ and symmetric interval $[-a,a]$: ${\int }_{-a}^{a}f=0$.

$\bullet$ Apply the quotient rule:

$${\log }_2\frac{x+y}{x-y}=2\Rightarrow \frac{x+y}{x-y}=4$$

$\bullet$ Substitute $x+y=8$:

$$\frac{8}{x-y}=4\Rightarrow x-y=2$$

$\bullet$ Solve the system:

$$x=5,y=3$$

$\bullet$ Evaluate:

$$2^x-2^y=32-8=24$$

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{\sin 2x}{x}}=2\cdot \underset{x\to 0}{\lim }{\displaystyle \frac{\sin 2x}{2x}}=2\cdot 1=2}$. So $a=2$.

${\displaystyle \frac{2-3\mathrm{/}x+1\mathrm{/}{x}^2}{1+5\mathrm{/}{x}^2}}\to {\displaystyle \frac{2}{1}}=2$.

Combinăm: ${\log }_3(x(x-6))=3\Rightarrow x^2-6x=27\Rightarrow x^2-6x-27=0$. Rădăcinile sunt $x=9$ și $x=-3$. Domeniul cere $x>6$, deci $x=9$ (cealaltă este străină).

With $a=2x$ and $b=3$: $T_{k+1}=\left(\genfrac{}{}{0px}{}{n}{k}\right)(2x{)}^{n-k}{3}^k$.

$\bullet$ Evaluate each special entry:

$$i^2=-1,\lg 100=2,\ln 1=0$$

$$\sin \frac{\pi }{2}=1,i^{24}=1,C_4^3=4$$

$\bullet$ Assemble the numerical matrix:

$$A=\left(\begin{array}{cccc}1& 0& 2& 4\\ -1& 1& 3& 2\\ 0& 1& 3& 0\\ 1& 1& 4& 7\end{array}\right)$$

$\bullet$ Row-reduce. $R_2+=R_1$, $R_4-=R_1$:

$$\left(\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& 5& 6\\ 0& 1& 3& 0\\ 0& 1& 2& 3\end{array}\right)$$

$\bullet$ $R_3-=R_2$, $R_4-=R_2$:

$$\left(\begin{array}{cccc}1& 0& 2& 4\\ 0& 1& 5& 6\\ 0& 0& -2& -6\\ 0& 0& -3& -3\end{array}\right)$$

$\bullet$ $R_4-=\frac{3}{2}{R}_3$ gives upper-triangular with pivots $1,1,-2,6$:

$$\det A=1\cdot 1\cdot (-2)\cdot 6=-12$$

$\int \text{\hspace{0.17em}⁣}({\displaystyle \frac{1}{x}}-{\displaystyle \frac{1}{x+1}})dx=\ln \mathrm{\mid }x\mathrm{\mid }-\ln \mathrm{\mid }x+1\mathrm{\mid }+C=\ln \text{\hspace{0.17em}⁣}\mid {\displaystyle \frac{x}{x+1}}\mid +C$.

$g(0)=1>0$, $g(\pi \mathrm{/}2)=0-\pi \mathrm{/}2<0$. Since $g$ is continuous, by the IVT there is $c\in (0,\pi \mathrm{/}2)$ with $g(c)=0$, i.e. $\cos c=c$.