Can we evaluate these log expressions without using a calculator?

Can we evaluate these log expressions without using a calculator?

Question

1. Without using tables [or calculators], find the value of

$$\begin{align*} 2 \lg 5 - \frac{1}{2} &\lg 9 + \lg 0.12.\\ [\lg x &= \log_{10} x] \end{align*}$$

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1. If $\lg 2 = 0.30103$ and $\lg 3 = 0.47712$, find, without using tables [or calculators], the values of $\lg 6$ and $\lg 15$.

Suggestion

If we can’t use calculators, we have to simplify this expression. So what rules and identities do we know for logarithms?

How could we rewrite the required logarithms in terms of the logarithms we are given? Remember that all logarithms are to base 10!

Solution

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

$$\begin{align*} \lg 25+\lg \frac{1}{3}+\lg \frac{12}{10} &= \lg \left( \frac{25\times12}{3\times10}\right)\\ &=\lg \left( 10\right)\\ &=1. \end{align*}$$

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

$$\begin{align*} \lg 15 &= \lg \frac{10\times3}{2}\\ &= \left( \lg 10 \right) + \left( \lg 3 \right) - \left( \lg 2 \right)\\ &\approx 1 + 0.47712 - 0.30103\\ &= 1.1761. \end{align*}$$