solve: $log_2(3x + 1)=4$\na $\frac{17}{3}$\nb 5\nc $\frac{7}{3}$\nd 9

Explanation:

Step1: Convert to exponential form

By the definition of logarithms, if $\log_{a}b = c$, then $b=a^{c}$. So, $\log_{2}(3x + 1)=4$ can be rewritten as $3x+1 = 2^{4}$.

Step2: Calculate the value of $2^{4}$

$2^{4}=16$, so the equation becomes $3x + 1=16$.

Step3: Solve for $x$

Subtract 1 from both sides: $3x=16 - 1=15$. Then divide both sides by 3: $x=\frac{15}{3}=5$.

Answer:

B. 5