Daily math problems for August 11, 2026 — Linear Function, Complex Numbers, Functions & more

$3x = 14 - 5 = 9 \Rightarrow x = 3$.

$(1 + i)^8 = \!\left[\sqrt{2}\right]^{8} \cdot \!\left(\cos 2\pi + i\sin 2\pi\right) = 16 \cdot (1 + 0) = 16$.

$f$ is invertible iff its slope $a - 2 \ne 0$. So $f$ fails to be invertible exactly when $a = 2$ (it then becomes the constant $5$, neither injective nor surjective).

Using $1-\cos x = 2\sin^2(x/2)$: $\displaystyle\lim_{x\to 0}\dfrac{2\sin^2(x/2)}{x^2} = \dfrac{1}{2}\lim_{x\to 0}\left(\dfrac{\sin(x/2)}{x/2}\right)^2 = \dfrac{1}{2}\cdot 1 = \dfrac{1}{2}$.

$\bullet$ The astroid meets the axes at $(\pm a, 0)$ and $(0, \pm a)$, so it fits in a square of side $2a$.

$\bullet$ Given side $= 4$:

$$2a = 4 \;\Rightarrow\; a = 2$$

$\bullet$ Apply the astroid area formula:

$$\mathcal{A} = \frac{3\pi a^2}{8} = \frac{3\pi \cdot 4}{8} = \frac{3\pi}{2}$$

$4p = 8 \Rightarrow p = 2$. Focus: $(2, 0)$.

The order of $g$ is the smallest positive integer $n$ such that $g^n = e$ (or infinity if no such $n$ exists).

$\vec{u} \cdot \vec{v} = a + (-2) = a - 2 = 0 \Rightarrow a = 2$.

$\bullet$ Simplify each logarithm:

$$\log_{\sqrt{2}}\!\left(\tfrac{1}{x-3}\right) = -2\log_2(x-3)$$

$$\log_{100}(10^x) = \tfrac{x}{2}$$

$\bullet$ Combine the numerator:

$$\log_2(x+25) - 2\log_2(x-3) = \log_2\frac{x+25}{(x-3)^2}$$

$\bullet$ The equation becomes:

$$\frac{\log_2\frac{x+25}{(x-3)^2}}{x/2} = \frac{2}{7}$$

$$\log_2\frac{x+25}{(x-3)^2} = \frac{x}{7}$$

$\bullet$ Test $x = 7$:

$$\log_2\frac{32}{16} = \log_2 2 = 1 = \frac{7}{7}$$

$10^3 - 10 = 990$, and $990 / 6 = 165$. The induction proof factors $n^3 - n = (n-1)n(n+1)$, which is the product of three consecutive integers and is therefore divisible by both $2$ and $3$, hence by $6$.