Before you get started, take this readiness quiz.
1. Simplify \(\left(\sqrt\right)^\).
2. Evaluate \(2^3\).
3. Evaluate \(3^\).
4. Write \(\sqrt[3] \) using exponents.
It works well to ‘undo’ an operation with another operation. Subtracting ‘undoes’ addition, multiplication ‘undoes’ division, taking the square root ‘undoes’ squaring.
Subtraction 'undoes' addition:
What undoes the exponential \(e^x\)?
and more generally, \(a^x\)?
In the previous section we learned about exponential expressions and how to evaluate them. In this section we will learn about a related type of expression, namely, the logarithmic expression.
Suppose we want to solve a simple exponential equation \(2^x=16\). Because we know how to evaluate it, we can use trial and error with a calculator to find an approximation to the value of \(x\). What we are looking for is the value of the exponent we need to raise the base 2 to to arrive at 16. This exponent (which depends on the base and the result) is called \(\log_2(16)\).
Assume \(b>0\). Then \(\log_b(a)\) ('the logarithm with base \(b\) of a') is the exponent we need on the base \(b\) to arrive at a value of \(a\),
or in other words, \(\log_b(a)\) is the solution to
You might wonder why we assume the base of the logarithm should be positive. What are the issues when this isn't the case?
Evaluate \(\log_5 25\).
We are looking here for the exponent we need to raise \(5\) to in order to arrive at 25. But we know \(5^2=25\). It follows that
Now, we will also note that
since we need to raise the base \(b\) to the exponent \(c\) in order to arrive at \(b^c\).
Evaluate \(\log_5 (25\sqrt)\).
We are looking here for the exponent we need to raise \(5\) to in order to arrive at \(25\sqrt\). But we know \(25\sqrt=5^2\cdot 5^\frac12=5^=5^\frac52\). It follows that
Note that just solving equations \(x^2=15\) involves the use of the square root, solving an equation \(2^x=15\) involves the logarithm. Most calculators reflect this analogy by the placement of the buttons. There are typically two easily accessible logarithm buttons: 'log' which is shorthand for \(\log_\) and 'ln' which is shorthand for \(\log_e\). Others logarithms will require additional understanding or a fancy calculator to approximate. Here we will give some examples using only the trial and error method for evaluation.
Approximate \(\log_5 7\).
This is equivalent to finding an approximate solution to \(5^x=7\).
We note that \(5^1=5\) and \(5^2=25\), so the exponent we seek is between these to exponents since \(5\), \(5^\) and \(5^\). we find that our desired exponent is between \(1.2\) and \(1.3\) and probably closer to 1.2. Evaluating \(5^\approx 7.01\) is enough to know the exponent we seek is between \(1.2\) and \(1.21\), probably closer to the latter. Evaluating \(5^\approx 6.9992\) is enough to conclude that our exponent is between \(1.209\) and \(1.21\) and, in particular, rounded to the nearest hundredth,
\(\log_5 (7)\approx 1.21.\)
Approximate \(\log_>(7)\) to the nearest hundredth
Answer
The solution to the equation \(x=a^y\) is written
So the chart becomes:
Subtraction 'undoes' addition:
