Daily Math · 2026-05-31

Question 1

Rezolvați $5^{x+1} = 125$.

Accepted Answer

Avem $125 = 5^3$, deci $5^{x+1} = 5^3 \Rightarrow x + 1 = 3 \Rightarrow x = 2$.

Answer

$x = 1$

Answer

$x = 2$

Answer

$x = 3$

Answer

$x = 4$

Question 2

In an arithmetic progression, $a_3 = 11$ and $a_7 = 27$. Determine $a_5$.

Accepted Answer

$a_5 = \dfrac{a_3 + a_7}{2} = \dfrac{11 + 27}{2} = 19$.

Answer

$15$

Answer

$17$

Answer

$19$

Answer

$22$

Question 3

Let $f(x)=C_x^3+\log_2(x+1)$ and $A=\begin{pmatrix}f(3)&	an\frac{\pi}{4}&2^{\lim_{x	o\infty}\frac{x^{23}}{x!}}\ \frac{f(7)}{38}&\log_{\sqrt{3}}f(3)&i^4\ \lg 10^{-48}&P_4&\ln 1\end{pmatrix}$. Find $\det(A)$.

Accepted Answer

$\bullet$ 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:

$$	an	frac{\pi}{4} = 1$$

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

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

$\bullet$ Assemble the matrix:

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

$\bullet$ Expand along row 1:

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

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

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

Answer

$0$

Answer

$24$

Answer

$-72$

Answer

$48$

Answer

$-24$

Answer

$72$

Problems & worked solutions

Problem 1 — Logarithms

Problem 2 — Arithmetic Sequences

Problem 3 — Matrices