Evaluating Logarithmic Functions at Given x-Values
What Logarithms Mean and How They Work
In other words, when given an expression like
$$\log_{a}{(y)}$$
this is asking “\(a\) to the what power gives you \(y\)?” Note that the subscript of the logarithm \(a\) is often referred to as the base of the logarithm, while the stuff of the inside of the log in parentheses is often called the argument of the log. Below we practice evaluating logs
\(\log_{2}{(4)}\) is asking “\(2\) to what power will give you \(4\)?”
The answer in this case is \(2\), so we write \(\log_{2}{(4)}=2\)
\(\log_{3}{(81)}\) is asking “\(3\) to what power will give you \(81\)?”
The answer here is \(4\) because \(3^4=3^2\cdot 3^2=9\cdot 9=81\). So we write \(\log_{3}{(81)}=4\).
\(\log_{5}{(5)}\) is asking “\(5\) to what power gives you \(5\).” This is only tricky if you overthink it.
\(\log_{5}{(5)}=1\) because \(5^1=5\).
This question is asking “\(2\) to what power will give us \(\frac{1}{16}\)?” or, in other words, “what exponent \(n\) do I need so that \(2^n=\frac{1}{16}\)?”
Note that if we raise \(2\) to the \(4\)th power, we get \(2^4=16\), but we want to raise \(2\) to some power that reciprocates this, putting the \(16\) in the denominator of the fraction. A power of \(4\) gives us \(2^4=16\), but a power of \(-4\) gives us \(2^{-4}=\frac{1}{16}\) (because negative powers flip/reciprocate fractions)
Logs as Functions
Given that we can evaluate logs like \(\log_{2}{(2)}\), \(\log_{2}{(4)}\), \(\log_{2}{(16)}\), \log_{2}{\left(\frac{1}{4}\right)}, i.e. where all the logs have the same base, but differ in their arguments, it makes sense that we could treat logs as functions.
Nothing really new to see in the above definition; we are merely defining logarithmic functions precisely. How they are evaluated is essentially the same as what we did in the examples above.
Note that \(f(9)=\log_{3}{(9)}\), which asks “\(3\) to what power gives you \(9\)?” The answer is obviously \(2\), so we write \(f(9)=\log_{3}{(9)}=2\).
\(f(27)=\log_{3}{(27)}=3\) for similar reasons (noting that \(3^3=27\)).
\(f(3)=\log_3(3)=1\) because raising \(3\) to the \(1\)st power gives you \(3\).
Now, for \(\log_{3}(3.45)\), notice that \(3.45\) is only marginally greater than \(3\), so one might suspect that the power \(n\) needed so that \93^n=3.45\) would be slightly bigger than 1. After a few guesses, plugging different values into a calculator, we see that \(n=1.127\) does the trick (it’s a suitable approximation).
Note that in the last example, we plugged a decimal number (namely \(3.45\)) into a logarithm, and got a decimal value as an output. This demonstrates that logarithmic functions are not just defined on fractions and whole numbers. It turns out that logarithmic functions of the form \(\log_{a}(x)\) where \(a>0\)) are defined for any positive value \(x\). In other words, you can plug in any decimal number greater than 0 in and expect to get an output. Might not be a pretty output value, but it still exists!
A Note on Natural Logs and Logs with Commonly Used Bases
Some logs with certain bases are used so commonly that mathematicians have gotten lazy about writing them the way you see demonstrated above. Below are a few commonly used logarithmic functions with their unique notations.
Log base 10
\(log(x)=\log_{10}{(x)}\)
Basically, if we omit the base in our notation, we assume that log base 10 is what is being used.
Log base \(e=2.71828…\)
\(\ln(x)=\log_{e}{(x)}\)
Log base 2
\(\lg(x)=\log_{2}{(x)}\)
\(f(2)=1\)
\(f(4)=2\)
\(f\left(\frac{1}{2}\right)=-1\)
\(f\left(\frac{1}{4}\right)=-2\)
\(f(2)=\log_2(2)\) asks “\(2\) to what power gives \(2\)?” Or, in other words, “what should the exponent be in the equation \(2^{?}=2\).” Certainly, the answer is \(1\).
\(f(4)=\log_2(4)\) asks “\(2\) to what power gives \(4\)?” Or, in other words, “what should the exponent be in the equation \(2^{?}=4\).” The answer is \(2\).
\(f\left(\frac{1}{2}\right)=\log_{2}\left(\frac{1}{2}\right)\) asks “\(2\) to what power gives \(\frac{1}{2}\)?” Or, in other words, “what should the exponent be in the equation \(2^{?}=\frac{1}{2}\).” The answer is \(-1\). (Because \(2^{-1}=\left(\frac{2}{1}\right)^{-1}=\left(\frac{1}{2}\right)^1=\frac{1}{2}\))
\(f\left(\frac{1}{4}\right)= \log_{2}\left(\frac{1}{4}\right) \) asks “\(2\) to what power gives \(\frac{1}{4}\)?” Or, in other words, “what should the exponent be in the equation \(2^{?}=\frac{1}{4}\).” The answer is \(-2\). (Because \(2^{-2}=\left(\frac{2}{1}\right)^{-2}=\left(\frac{1}{2}\right)^2=\frac{1}{4}\))
\(f\left(\frac{1}{81}\right)=4\)
\(f(1)=0\)
\(f(3)=-1\)
\(f(81)=-4\)
\(f\left(\frac{1}{81}\right) =\log_{\frac{1}{3}}\left(\frac{1}{81}\right)\) asks “\(\frac{1}{3}\) to what power gives \(\frac{1}{81}\)?” Or, in other words, “what should the exponent be in the equation \(\left(\frac{1}{3}\right)^{?}=\frac{1}{81}\).” The answer is \(4\).
\(f(1)=\log_{\frac{1}{3}}(1)\) asks “\(\frac{1}{3}\) to what power gives \(1\)?” Or, in other words, “what should the exponent be in the equation \(\left(\frac{1}{3}\right)^{?}=1\).” The answer is \(0\) because anything to the power \(0\) is \(0\) (however, \(0^0\) is undefined).
\(f(3)= \log_{\frac{1}{3}}(3)\) asks “\(\frac{1}{3}\) to what power gives \(3\)?” Or, in other words, “what should the exponent be in the equation \(\left(\frac{1}{3}\right)^{?}=3\).” The answer is \(-1\). (Because \(\left(\frac{1}{3}\right)^{-1}=3^1=3\))
\(f(81)= \log_{\frac{1}{3}}(81)\) asks “\(\frac{1}{3}\) to what power gives \(81\)?” Or, in other words, “what should the exponent be in the equation \(\left(\frac{1}{3}\right)^{?}=81\).” The answer is \(-4\). (Because \(\left(\frac{1}{3}\right)^{-4}=3^4=81\))