## Intermediate Algebra: Connecting Concepts through Application

$x = -\frac{5}{2}$ The student added $3 + 5$ together to obtain $8$. This is where Frank should have treated $\log_2 (x+3)$ as separate from $5$. He should not have added them together. Instead, Frank should have subtracted $5$ from both sides. (Refer to correct working).
$\log_2 (x+3) + 5 = 4$ $\log_2 (x+3) = 4 -5$ $\log_2 (x+3) = -1$ $2^{-1} = x+3$ $0.5 - 3 = x$ $x = -\frac{5}{2}$ The student added $3 + 5$ together to obtain $8$. This is where Frank should have treated $\log_2 (x+3)$ as separate from $5$. He should not have added them together. Instead, Frank should have subtracted $5$ from both sides (Refer to correct working).