Review question

# Can we evaluate these log expressions without using a calculator? Add to your resource collection Remove from your resource collection Add notes to this resource View your notes for this resource

Ref: R9976

## Solution

1. 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*}
1. 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*}