Powers, Radicals, Logarithms practice questions — AP Calculus BC

$\dfrac{1}{\sqrt{3} - 1} \cdot \dfrac{\sqrt{3} + 1}{\sqrt{3} + 1} = \dfrac{\sqrt{3} + 1}{3 - 1} = \dfrac{\sqrt{3} + 1}{2}$.

$\log_2 6 + \log_2 \dfrac{8}{3} = \log_2\!\left(6 \cdot \dfrac{8}{3}\right) = \log_2 16 = 4$.

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

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

Apply the power law to the first term. $$2\log_5 10 = \log_5 10^2 = \log_5 100$$

Now use the subtraction (quotient) law. $$\log_5 100 - \log_5 4 = \log_5 \frac{100}{4} = \log_5 25$$

Finally write $25$ as a power of $5$. $$\log_5 25 = \log_5 5^2 = 2$$

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

Use the product rule for logarithms: $$\log_3(9x)=\log_3 9+\log_3 x$$ Since $9=3^2$, we have $\log_3 9 = 2$, and we are given $\log_3 x = 4$: $$\log_3(9x)=2+4=6$$

Apply the quotient rule to merge the difference into a single logarithm: $$\log_2 80 - \log_2 5 = \log_2 \frac{80}{5}$$ Simplify the argument and evaluate, since $2^4 = 16$: $$\log_2 16 = 4$$

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

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