Powers, Radicals, Logarithms

$\log_2(x(x-2)) = 3$ → $x^2 - 2x = 8$ → $x^2 - 2x - 8 = 0$ → $(x-4)(x+2)=0$. Since $x > 2$, $x = 4$.

$\log_3(81) = \log_3(3^4) = 4$ and $\log_3(3) = 1$. The difference is $4 - 1 = 3$.

$2^5 = 32$, so $\log_2 32 = 5$.

$\sqrt{12 \cdot 3} = \sqrt{36} = 6$.

$\sqrt[3]{8} = 2$, so $8^{2/3} = 2^2 = 4$.

$\sqrt{50} = \sqrt{25 \cdot 2} = \sqrt{25} \cdot \sqrt{2} = 5\sqrt{2}$.

$8^{2/3} = \left(8^{1/3}\right)^{2} = 2^{2} = 4$.

(Or: $8^{2} = 64$, then $\sqrt[3]{64} = 4$.)

$\log_4 8 = \dfrac{\log_2 8}{\log_2 4} = \dfrac{3}{2}$.

$27 = 3^3$, so $3^{x+1} = 3^3 \Rightarrow x + 1 = 3 \Rightarrow x = 2$.

$2^x < 2^3 \Leftrightarrow x < 3$, i.e. $x \in (-\infty, 3)$.