Daily math problems for August 13, 2026 — Functions, Logarithms, Matrices & more

We need $x - 2 \ge 0$, i.e. $x \ge 2$. The domain is $[2, +\infty)$.

By the chain rule for logs, $\log_a c = \log_a b \cdot \log_b c = 2 \cdot 3 = 6$.

$\operatorname{tr}(A) = 2 + 3 + 5 = 10$.

By Fermat, $3^{6} \equiv 1 \pmod 7$, so $\hat{3}^{\,6} = \hat{1}$.

$\int \ln x \, dx = x \ln x - x + C$. Evaluate: $(e \cdot 1 - e) - (1 \cdot 0 - 1) = 0 - (-1) = 1$.

$3 \cdot (-5) = (m - 1)\,(-(m+1))$, i.e. $-15 = -(m-1)(m+1) = 1 - m^2$, so $m^2 = 16$ and $m \in \{-4, 4\}$.

$\mathbb{Q}$ has all the field axioms — every non-zero rational has a rational reciprocal. It's a field.

$\bullet$ Rewrite in power form using $\ln\sqrt[3]{x^{10}}=\tfrac{10}{3}\ln x$:

$$f(x)=x^{-7/3}+\tfrac{10}{3}\ln x$$

$\bullet$ Differentiate once:

$$f'(x)=-\tfrac{7}{3}x^{-10/3}+\tfrac{10}{3x}$$

$\bullet$ Differentiate again:

$$f''(x)=\tfrac{70}{9}x^{-13/3}-\tfrac{10}{3x^2}$$

$\bullet$ Plug in $x=2$ in $f'$ and $x=3$ in $f''$:

$$f'(2)=-\tfrac{7}{3\cdot 2^{10/3}}+\tfrac{5}{3}$$

$$f''(3)=\tfrac{70}{9\cdot 3^{13/3}}-\tfrac{10}{27}$$

$V = \pi \int_0^1 (x^2 - x^4) \, dx = \pi\!\left[\dfrac{x^3}{3} - \dfrac{x^5}{5}\right]_0^1 = \pi\!\left(\dfrac{1}{3} - \dfrac{1}{5}\right) = \dfrac{2\pi}{15}$.

$\bullet$ Evaluate the function values:

$$f(3) = C_3^3 + \log_2 4 = 1 + 2 = 3$$

$$f(7) = C_7^3 + \log_2 8 = 35 + 3 = 38$$

$\bullet$ So $f(7)/38 = 1$ and $\log_{\sqrt{3}} f(3) = \log_{\sqrt{3}} 3 = 2$.

$\bullet$ Reduce the trig / limit / log / factorial entries:

$$\tan\tfrac{\pi}{4} = 1$$

$$2^{\lim x^{23}/x!} = 2^0 = 1$$

$$i^4 = 1,\;\; \lg 10^{-48} = -48,\;\; P_4 = 4! = 24,\;\; \ln 1 = 0$$

$\bullet$ Assemble the matrix:

$$A = \begin{pmatrix}3 & 1 & 1\\ 1 & 2 & 1\\ -48 & 24 & 0\end{pmatrix}$$

$\bullet$ Expand along row 1:

$$\det A = 3(2\cdot 0 - 1\cdot 24) - 1(1\cdot 0 - 1\cdot(-48)) + 1(1\cdot 24 - 2\cdot(-48))$$

$$\det A = 3(-24) - 1(48) + 1(120)$$

$$\det A = -72 - 48 + 120 = 0$$