Why are these equations with logs equivalent?

Question

I have two equations:

$$\frac{16}{3}(ln4 - \frac{2}{3}) + \frac{4}{9}$$

and

$$\frac{4}{9}(4ln(64) - 7)$$

Why are they equivalent?

So I know $2log_2{8} = log_2{8^2}$$

Answer 1

$$ \frac{4}{9}(4\ln64-7) = \frac{16}{9}\ln4^3-\frac{28}{9} = \frac{16}{9} \cdot 3\ln 4 - \frac{28}{9} = \frac{16}{3}\ln4 - \frac{32}{9} + \frac{4}{9} =\frac{16}{3}\ln4 - \frac{16}{3}\cdot \frac{2}{3} + \frac{4}{9}$$