Can we use the laws of logs to find $x$ and $y$?

Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

We will simplify the right hand side of the above equation as far as possible. By using the logarithm quotient rule we get \[\begin{align*} \log{\frac{4}{9}} - \log{\frac{9}{125}} + 2\log{\frac{9}{5}} & = \log{4} - \log{9} - \log{9} + \log{125} + 2\log{9} - 2\log{5} \\ & = \log{4} + \log{125} - 2\log{5}. \end{align*}\]

Now as \(125 = 5^3\) we can use the logarithm power rule to get \[\log{4} + \log{5^3} - 2\log{5} = \log{4} + 3\log{5} - 2\log{5} = \log{4}+\log{5}=\log{20}.\] Therefore, we get \(\log{x} = \log{20}\) and thus \(x = 20\).

Now for \(y\) we get the equation \(2 + \log{y} = \log{20}\). As all logarithms are to base \(10\) we can write \(2 = \log{100}\) and therefore \[\begin{align*} \log{y} & = \log{20} - \log{100} \\ & = \log{\frac{20}{100}} \\ & = \log{\frac{1}{5}}. \end{align*}\]

Hence \(y = \frac{1}{5}\).