​ please help! An explanation would be helpful :D

$\log_{b} \left(\dfrac 2 3\right) + \log_{b} \left(\dfrac 3 2\right) = 1$

Let $x = \log_b ({2 \over 3})$ and $y = \log_b ({2 \over 3})$

Then $b^x = {2 \over 3}$ and $b^y = {3 \over 2}$

Multiplying the two gives us $b^x \times b^y = b^{x + y} = 1$

Now, notice that the only way the expression equals 1 is if $x + y = \color{brown}\boxed{0}$.

Note: In general, $\log_b({x}) + \log_b({y}) = \log_b({xy})$