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 $8^{2\mathrm{/}3}=2^2=4$.

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

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

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

${\log }_{4}8={\displaystyle \frac{{\log }_{2}8}{{\log }_{2}4}}={\displaystyle \frac{3}{2}}$.

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

$2^x<2^3\iff x<3$, i.e. $x\in (-\mathrm{\infty },3)$.