Review question

# 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

Ref: R7471

## Solution

If $\log{x} = 2 + \log{y} = \log{\frac{4}{9}} - \log{\frac{9}{125}} + 2\log{\frac{9}{5}},$ where all logarithms are to base $10$, find $x$ and $y$ without using tables.

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 $y = \frac{1}{5}$.