Let b > 1. Find the value of
\(\log_{b} \left(\dfrac 2 3\right) + \log_{b} \left(\dfrac 3 2\right).\)
+2666 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, \(\log_b({x}) + \log_b({y}) = \log_b({xy})\)
