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:

$$x = 5,\quad y = 3$$

$\bullet$ Evaluate:

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

$\displaystyle\lim_{x \to 0} \dfrac{\sin 2x}{x} = 2 \cdot \lim_{x \to 0} \dfrac{\sin 2x}{2x} = 2 \cdot 1 = 2$. So $a = 2$.

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

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 $b = 3$: $T_{k+1} = \binom{n}{k}(2x)^{n-k} 3^k$.

$\bullet$ Evaluate each special entry:

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

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

$\bullet$ Assemble the numerical matrix:

$$A = \begin{pmatrix}1 & 0 & 2 & 4\\ -1 & 1 & 3 & 2\\ 0 & 1 & 3 & 0\\ 1 & 1 & 4 & 7\end{pmatrix}$$

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

$$\begin{pmatrix}1 & 0 & 2 & 4\\ 0 & 1 & 5 & 6\\ 0 & 1 & 3 & 0\\ 0 & 1 & 2 & 3\end{pmatrix}$$

$\bullet$ $R_3 \mathrel{-}= R_2$, $R_4 \mathrel{-}= R_2$:

$$\begin{pmatrix}1 & 0 & 2 & 4\\ 0 & 1 & 5 & 6\\ 0 & 0 & -2 & -6\\ 0 & 0 & -3 & -3\end{pmatrix}$$

$\bullet$ $R_4 \mathrel{-}= \tfrac{3}{2} R_3$ gives upper-triangular with pivots $1, 1, -2, 6$:

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

$\int \!\left(\dfrac{1}{x} - \dfrac{1}{x+1}\right) dx = \ln|x| - \ln|x + 1| + C = \ln\!\left|\dfrac{x}{x + 1}\right| + C$.

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