$\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$ Show $\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$
from LS
$\displaystyle 1+ \frac{\log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2(3/2) + \log_2(2/3(n+1))}{\log_2(3/2)} = \frac{\log_2\big(\frac{6(n+1)}{6}\big)}{\log_2(3/2)} = \frac{\log_2(n+1)}{\log_2(3/2)}$
is this right?

Yes, your solution is correct, but you could have done it a more "natural" way. In this case of your problem, there two useful properties of logarithms that you could use: exponents can be brought out front and multiplication under the logarithm sign turns into addition when the multiplicands are separated.
1)$$
\log_{a}{\frac{x}{y}}=\log_{a}{\left[\left(\frac{y}{x}\right)^{-1}\right]}=
-1\cdot\log_{a}{\frac{y}{x}}=-\log_{a}{\frac{y}{x}}.
$$
2)$$
\log_{a}\left({xy}\right)=\log_{a}{x}+\log_{y}.
$$

$$
1+\frac{\log_2\left[\frac{2}{3}(n+1)\right]}{\log_2\frac{3}{2}} =
1+\frac{\log_2\frac{2}{3}+\log_2\left(n+1\right)}{\log_2\frac{3}{2}} =\\
1+\frac{-\log_2\frac{3}{2}+\log_2\left(n+1\right)}{\log_2\frac{3}{2}}=
1-\frac{\log_2\frac{3}{2}}{\log_2\frac{3}{2}}+\frac{\log_2\left(n+1\right)}{\log_2\frac{3}{2}}=\\
1-1+\frac{\log_2\left(n+1\right)}{\log_2\frac{3}{2}}=
\frac{\log_2\left(n+1\right)}{\log_2\frac{3}{2}}
$$
