When working with exponents, we employ a variety of properties and laws to help simplify and evaluate exponential expressions.

Common Logarithms

Common logarithms were the first logarithms introduced to carry out complicated calculations in the decimal number system (base \(10\)).

The log function

log

on your calculator works in base \(10\).

Evaluate.

a. \(\log (0.001)\)

b. \(\log(70)\)

c. \(\log (-100)\)

Solution

a. Evaluating this logarithm, we have

You can also use the log function on your calculator log. This works in base \(10\) (common base).

b. \(\log(70) = 1.84509804 \ldots\)

Since this answer is the value of an exponent, it is important to keep \(3\) or \(4\) decimal places in your answer when rounding.

Therefore, \(\log (70) \approx 1.845\). This means that \(10^{1.845} \approx 70\).

Evaluate.

a. \(\log (0.001)\)

b. \(\log(70)\)

c. \(\log (-100)\)

Solution

c.The calculator returns an error when asked to determine \(\log (-100)\).

Why is this?

If we let \(\log (-100)= n\), then \(10^n = -100\).

There is no value for \(n\) such that \(10^n\) produces a negative value since the base is positive.

Also, remember that the domain of the logarithmic function \(y = \log_{c}(x), c \gt 0,~ c \neq 1\) is \(\left\{ x \mid x \gt 0, x \in \mathbb{R} \right\}\).

Evaluate.

a. \(4^{\log_{4}{(64)}}\qquad\qquad\)b. \(5^{\log_{5} {\left(\frac{1}{25}\right)}} \qquad\qquad\)c. \(3^{\log_{3}{(20)}}\)

Solution

Notice \(4^{\log_{4}{\color{NavyBlue}(64)}} = {\color{NavyBlue}64}\) and \(5^{\log_{5}{ {\color{NavyBlue} \left( \frac{1}{25} \right)}}} = {\color{NavyBlue}{\dfrac{1}{25}}}\). It would seem that \(3^{\log_{3}(20)}\) should equal \(20\).

c. If we let \({\color{BrickRed}3}^{\color{Mulberry}{\log_{3}(20)}} = {\color{NavyBlue}n}\) and convert to logarithmic form, we have

Thus, \(3^{\log_{3}(20)} = 20\).