- Without using tables [or calculators], find the value of \[\begin{align*} 2 \lg 5 - \frac{1}{2} &\lg 9 + \lg 0.12.\\ [\lg x &= \log_{10} x] \end{align*}\]
Using the logarithm laws, we can rewrite the first two terms as
\[\begin{align*} 2 \lg 5 &= \lg 5^2 = \lg 25\\ - \frac{1}{2}\lg 9 &= \lg 9^{-\frac{1}{2}} = \lg \frac{1}{\sqrt{9}} = \lg \frac{1}{3}. \end{align*}\]Now we can evaluate the original expression
\[\begin{align*} \lg 25+\lg \frac{1}{3}+\lg \frac{12}{10} &= \lg \left( \frac{25\times12}{3\times10}\right)\\ &=\lg \left( 10\right)\\ &=1. \end{align*}\]- If \(\lg 2 = 0.30103\) and \(\lg 3 = 0.47712\), find, without using tables [or calculators], the values of \(\lg 6\) and \(\lg 15\).
Following the logarithm laws we can rewrite the logarithm of six as \[\lg 6 = \lg 2\times3 = \left(\lg 2\right)+\left(\lg 3\right).\]
Inserting the given values for \(\lg 2\) and \(\lg 3\) we therefore obtain \[\lg 6 = 0.30103 + 0.47712 \approx 0.7782.\]
Since both given values have a precision of 5 decimal places, we can approximate the solution to 4 decimal places.
Fifteen can be expressed as \[15=\frac{10\times3}{2}.\]
Since we know the base-10 logarithm of 2, 3 and 10 we can write
\[\begin{align*} \lg 15 &= \lg \frac{10\times3}{2}\\ &= \left( \lg 10 \right) + \left( \lg 3 \right) - \left( \lg 2 \right)\\ &\approx 1 + 0.47712 - 0.30103\\ &= 1.1761. \end{align*}\]
