If $\log 2 = .3010$ and $\log 3 = .4771$, the value of $x$ when $3^{x+3} = 135$ is approximately
$\textbf{(A) \ } 5  \qquad \textbf{(B) \ } 1.47 \qquad \textbf{(C) \ } 1.67 \qquad \textbf{(D) \ } 1.78 \qquad \textbf{(E) \ } 1.63$
Taking the logarithm in base $3$ of both sides, we get $x+3 = \log_3 135$. Using the property $\log ab = \log a + \log b$, we get $x+3 = \log_3 5 + \log_3 3^3$, or $x = \log_3 5$. Converting into base $10$ gives $x = \frac{\log 5}{\log 3} = \frac{1 - \log 2}{\log 3}$. Now, plugging in the values yeilds $\boxed{\textbf{(B) \ } 1.47  }$.