Step 1: Use the property of logarithms \( \frac{1}{\log_b x} = \log_x b \).
This simplifies the given equation to: \[ \log_x 2 + \log_x 3 + \log_x 4 + \log_x 5 + \log_x 6 = 1 \] Simplifying the sum gives: \[ \log_x (2 \times 3 \times 4 \times 5 \times 6) = \log_x 720 \] Thus, \( x = 720 \).
Draw a rough sketch for the curve $y = 2 + |x + 1|$. Using integration, find the area of the region bounded by the curve $y = 2 + |x + 1|$, $x = -4$, $x = 3$, and $y = 0$.
If the function \[ f(x) = \begin{cases} \frac{2}{x} \left( \sin(k_1 + 1)x + \sin(k_2 -1)x \right), & x<0 \\ 4, & x = 0 \\ \frac{2}{x} \log_e \left( \frac{2 + k_1 x}{2 + k_2 x} \right), & x>0 \end{cases} \] is continuous at \( x = 0 \), then \( k_1^2 + k_2^2 \) is equal to: