We need to find the value of the expression $6 + \log_b(\frac{1}{b^3}) + \log_b(\sqrt{b})$.

1. Problem Description

We need to find the value of the expression $6 + \log_b(\frac{1}{b^3}) + \log_b(\sqrt{b})$.

2. Solution Steps

First, let's simplify the logarithmic terms.

Recall that $\log_a(x^n) = n \log_a(x)$. Also $\frac{1}{x^n} = x^{-n}$.

$\log_b(\frac{1}{b^3}) = \log_b(b^{-3}) = -3\log_b(b)$.

Since $\log_b(b) = 1$, we have $\log_b(\frac{1}{b^3}) = -3$.

Next, let's simplify $\log_b(\sqrt{b})$.

$\sqrt{b} = b^{\frac{1}{2}}$.

So, $\log_b(\sqrt{b}) = \log_b(b^{\frac{1}{2}}) = \frac{1}{2}\log_b(b) = \frac{1}{2}$.

Now, substitute these simplified values back into the original expression:

$6 + \log_b(\frac{1}{b^3}) + \log_b(\sqrt{b}) = 6 + (-3) + \frac{1}{2} = 6 - 3 + \frac{1}{2} = 3 + \frac{1}{2} = \frac{6}{2} + \frac{1}{2} = \frac{7}{2}$.

3. Final Answer

$\frac{7}{2}$