Logarithms: What’s the Missing Exponent?
Your average school might not encounter logarithms until Algebra 2; however, there is nothing particularly more complex about “logs” than anything else in exponentiation, which we have already covered. Logs are merely asking the question “given some base, what is the exponent I need to apply to this base to return the value specified.” For example:
$$5^{?}=125$$
Well, yes, $5^3=125$ and so the answer is “three.” In mathematics, we represent this situation as
$$5^{?}=125\longrightarrow \log_{5}125=?$$
Namely, $\log_{5}125=3$
Herein, we will cover the basics of rooting as well as some of the fundamental laws and principles concerning these operations.
1. Logarithms: The Basics
If the rooting operation is asking “what is the base given an known exponent and known result” (i.e. $?^3=125$), and exponentiation is asking “what is the result if a exponentiate a quantity by some specific exponent” (i.e. $5^3=?$), then we can think of the relationship between exponentiation, rooting, and logarithms as tripartite with a direct “question” or “function” moving us between the different possible unknowns in any exponentiation system. Thus, for thee possible different quantitites $x$, $y$, and $z$, we can establish a function to resolve each if unknown.
- $x^y=?\longrightarrow$ translates as is given: $x^y=?$
- $?^y=z\longrightarrow$ translates as: $\sqrt[y]{z}=?$
- $x^?=z\longrightarrow$ translates as: $\log_{x}z=?$$
By makeing these “question mark” statements into functions, we can isolate the “solution” (the answer to the “?”) one one side of the equals sign and more appropraitely manipulate the expression as desired.
More mathematically put, the logarithm with a positive base $b$ and a given positive value “$a$ returns the exponent $n$ that can be applied to the base $b$ to return the given value $a$. This is represented as
$\log_{b}a=n\qquad$ where $\qquad{b^n=a}\quad\quad$ and $b\gt0$ and $b\neq1$ and $a\gt0$
When the base is unspecified (as in $\log{a}=n$), it is assumed that the base of the logarithm is 10; this is called the “common log“.
Examples:
$log_{2}4=2$
$log_{3}81=4$
$log100=2$
$log_{\tfrac{1}{2}}\left(\frac{1}{16}\right)=4$
Later in our math (and science) instruction you will encounter the natural logarithm. The natural logarithm is precisely the same operation as the logarithm, expect the natural logarithm merely has the specific base of a special, irrational constant “$e$“ or “Euler’s Number.” This number appears in many natural phenomena, and so the natural logarithm is often useful in various scientific models of natural laws. The natural logarithm is written lightly differently from the other logarithms. Since it appears most frequently in nature, it has been given a more abbreviated form:
$$\ln{a}$$
This is read as “the natural logarithm of a.” It can be thought of as the same as
$$\ln{a}=\log_{e}a$$
Euler’s Number will appear in many of our advanced mathematics courses, and you will examine many of its properties later. However, it is worth noting its use in logarithms now. Euler’s Number is an irrational number and is approximately $e=2.718281828…\quad$;
2. The Laws of Logs
Since the logarithms are fundamentally related to exponentials (both exponents and roots/fractional exponents), they share related properties derivative of the Five Laws of Exponents.
Fundamental Properties:
There are a few trivial definitions that are worth noting and remembering becuase they can be powerful in the simplification of many problems. We will draw paralells to each exponent definition to justify these properties of logs. Of course, all these properties and laws are given in general form for all logarithms, and so the below hold true for all logs, including natural and common logs.
Zero Power Rule
$$\log_{x}1=0$$
Proof!
Exponent Form:
$$x^0=1\quad\text{for all}\quad{x}$$
Logarithmic Equivalent:
$$\log_{x}1=0$$
Since we have defined the exponentiation form as axiomatic (for purposes of continuity to be discussed later), we can say that the logarithmic form is merely derived from this axiom and so is true.
Exponential Identity
$$\log_{x}x=1$$
Proof!
Exponent Form:
$$x^1=x\quad\text{for all}\quad{x}$$
Logarithmic Equivalent:
$$\log_{x}x=1$$
Truth Statement
$$\log_{b}b^x=x$$
Proof!
This property is often included in lists, but is not neccessarily that interesting when one sees its exponential equivalent:
$$b^x=b^x\quad\text{for all}\quad{x}$$
This is merely a truth statement, like $1=1$ or $x= x$. However, when present in the logarithmic form, this general truth statement is a bit obfuscated. Thus, we list it as a note to keep in mind because this can be a powerful property to have for logarithmic simplification.
$$\log_{b}b^x=x$$
Log Power Rule
$$b^{\log_{b}x}=x$$
Proof!
Assume $\log_{b}x=n$
Then, $b^n=x$ by definition of logarithms.
Thus, we can say from this that
$$b^{\log_{b}x}=x$$
since, $\log_{b}x=n$, and so $b^{\log_{b}x}=b^n$, and we established that $b^n=x$, and so
$$b^{\log_{b}x}=b^n=x$$
Thus, we have proven that
$$b^{\log_{b}x}=x$$
The Five Laws of Logarithms:
Logarithmic Law of Products
This parallels the Product Rule of Exponents from “The Five Laws of Exponents“
$$\log_{b}(x\cdot{y})=\log_{b}x+\log_{b}y$$
Click to read proof
Let $m=\log_{b}x$ and $n=\log_{b}y$
Then
$$m=\log_{b}x\longrightarrow{x=b^m}$$
$$n=\log_{b}y\longrightarrow{y=b^n}$$
By the Product Rule of Exponentiation we can say that
$$x(y)=(b^m)(b^n)=b^{m+n}$$
Since we are looking for the $log_{b}(xy)$, we should take the log of both sides of the identity given above.
$$x(y)=b^{m+n}$$
$$\log_{b}(xy)=\log_{b}(b^{m+n})$$
We can now use the definition of logs given in the “Truth Statement” under the “Fundamental Properties of Logs” and conclude that
given $\log_{x}(x^y)=y\longrightarrow{x^y=x^y}$
then $\log_{b}(b^{m+n})=m+n$
And so,
$$\log_{b}(xy)=\log_{b}(b^{m+n})=m+n$$
Now we can substitute back in our original expression for $m$ and $n$.
$$\log_{b}(xy)=m+n$$
$$\log_{b}(xy)=\log_{b}x+\log_{b}y$$
Logarithmic Law of Quotients
This parallels the Quotient Rule of Exponents from “The Five Laws of Exponents“
$$\log_{b}(\frac{x}{y})=\log_{b}x-\log_{b}y$$
Click to read proof
Let $m=\log_{b}x$ and $n=\log_{b}y$
Then
$$m=\log_{b}x\longrightarrow{x=b^m}$$
$$n=\log_{b}y\longrightarrow{y=b^n}$$
By the Quotient Rule of Exponentiation we can say that
$$\frac{x}{y}=\frac{b^m}{b^n}=b^{m-n}$$
Since we are looking for the $log_{b}(xy)$, we should take the log of both sides of the identity given above.
$$\frac{x}{y}=b^{m-n}$$
$$\log_{b}(\frac{x}{y})=\log_{b}(b^{m-n})$$
We can now use the definition of logs given in the “Truth Statement” under the “Fundamental Properties of Logs” and conclude that
given $\log_{x}(x^y)=y\longrightarrow{x^y=x^y}$
then $\log_{b}(b^{m-n})=m-n$
And so,
$$\log_{b}(\frac{x}{y})=\log_{b}(b^{m-n})=m-n$$
Now we can substitute back in our original expression for $m$ and $n$.
$$\log_{b}(\frac{x}{y})=m-n$$
$$\log_{b}(\frac{x}{y})=\log_{b}x-\log_{b}y$$
Logarithmic Law of Exponents
This parallels the Power of a Power Rule of Exponents from “The Five Laws of Exponents“
$$\log_{b}(x^k)=k\cdot\log_{b}x$$
Click to read proof
Let $m=\log_{b}x$
Then
$$m=\log_{b}x\longrightarrow{x=b^m}$$
By the Power of Powers Rule of Exponentiation we can say that
$$x=b^m$$
$$(x)^k=(b^m)^k$$
$$x^k=b^{mk}$$
Since we are looking for the $log_{b}(x^k)$, we should take the log of both sides of the identity given above.
$$x^k=b^{mk}$$
$$\log_{b}(x^k)=\log_{b}(b^{mk})$$
By the definition of logs in the Truth Statement given above in the “Fundamental Properties” of logs, we can say that since
$$\log_{r}(r^s)=s$$
then
$$\log_{b}(b^{mk}=mk$$
and so
$$\log_{b}(x^k)=\log_{b}(b^{mk})=mk$$
Recall that $m=\log_{b}x$, and so we can substitute back in this value for $m$ and return
$$\log_{b}(x^k)=mk$$
$$\log_{b}(x^k)=(\log_{b}x)(k)$$
By commutative property of multiplication, we can reorder
$$\log_{b}(x^k)=k\cdot\log_{b}x$$
Change of Logarithmic Base Law
Sometimes we need to change the base of our log to a different base. This can be done with this identity:
$$\log_{a}x=\frac{\log_{b}x}{\log_{b}a}$$
Click to read proof
Let $k=\log_{a}x$
Then
$$k=\log_{a}x\longrightarrow{x=a^k}$$
Now, let’s take the log of each side, but let’s also make this log have a different arbitrary base, $b$ (which is assumed $b>0$ and $b\neq1$
$$\log_{b}x=\log_{b}(a^k)$$
We can now use the Logarithmic Law of Exponents, as proven previously, to simplify this equation.
$$\log_{b}x=k\cdot\log_{b}(a)$$
Now we will slove for k by dividing both sides of the equation by $\log_{b}a$
$$\log_{b}x=k\cdot\log_{b}(a)$$
$$\frac{\log_{b}x}{\log_{b}(a)}=\frac{k\cdot\log_{b}a}{\log_{b}a}$$
$$\frac{\log_{b}x}{\log_{b}(a)}=k$$
Now, recall that $k=\log_{a}x$. We will substitute this back into our equation above.
$$\frac{\log_{b}x}{\log_{b}(a)}=\log_{a}x$$
or
$$\log_{a}x=\frac{\log_{b}x}{\log_{b}(a)}$$
Keep in mind that while the above properties help us greatly simplify many logarithmic problems, there are some logarithmic problems that cannot be simplified through any of these properties. For example these are NOT EQUAL
$$\log_{b}(x+y)\neq\log_{b}x+\log_{b}y$$
Also
$$\log_{b}(x-y)\neq\log_{b}x-\log_{b}y$$
Proof!
Assume $\log_{b}(x-y)=\log_{b}x-\log_{b}y$
Then, by the previously proven Logarithmic Law of Products (see above), we can say that
$$\log_{b}x-\log_{b}y=\log_{b}(xy)$$
And so
$$\log_{b}(x-y)=\log_{b}(xy)$$
Let’s now convert this statement into its exponential form
$$\log_{b}(x-y)=\log_{b}(xy)\longrightarrow{b^{\log_{b}(xy)}=x-y}$$
By the Log Power Rules proven previously (i.e. $b^{\log_{b}x}=x$), we know that $b^{\log_{b}(xy)=xy$, and so
$$b^{\log_{b}(xy)}=x-y\longrightarrow\text{implies}\longrightarrow{xy=x-y}$$
This is only true in limited cases where $x=\frac{x-y}{y}$ and not for all values. Thus, in general, we cannot assume that
$$\log_{b}(x-y)=\log_{b}x-\log_{b}y$$
And so we must generally assume that
$$\log_{b}(x-y)\neq\log_{b}x-\log_{b}y$$
3. Extending the Log Function: Negative Bases and $\log_{0}$
We will touch upon logarithms with negative bases; however, they are more the purview of Complex Analysis (a later topic in math), and so we will only present them and their general meaning. We should also talk about the $log_{0}$ for completion of the log operation.
$$\log_{0}x=y$$
The $\log_{0}$ is an odd question. Let’s consider it in its exponential form:
$$0^y=x$$
Well, raising zero to any power will always return zero, except in the caseof $0^0$, which is axiomatically defined to be $1$ for purposes of continuity (we will discuss continuity later). Thus,
$$0^y=0\,\text{for all}\,y\gt0$$
Keep in mind that $y$ cannot be negative, since
$$0^{-y}=\frac{1}{0^y)=\frac{1}{0}=\,\text{undefined for all}\,{y\gt0$$
Thus, $\log_{0}x=y$ works assuming $y$ is positive and $x=0$
The only other case where $\log_{0}x=y$ could be valid is in the case where $x=1$, in which case $y=0$, giving the axiom $\log_{0}1=0\longrightarrow{0^0=1}$.
Thus, the only cases where $\log_{0}x=y$ is a valid question are
$$\log_{0}0=x\,\text{for all}\,{x\gt0}\quad\quad\log_{0}1=0$$
$$\log_{b}x\,\text{for}\,x\lt0$$
You might have read in some introduction textbooks that “we can’t take the logarithm of zero or a negative number.” This this only partly true; while there are an infinitude of values in the Real Numbers for which we cannot do this, there are some select values in the Real Numbers for which this will work under certain constraints.
Let’s consider $log_{-x}y=z\longrightarrow{(-x)^z=y}$
We know from the section on exponentiation that raising a negative number to a power (negative or positive) has some fairly capricious behavior. For example, let’s consider some powers of $-2$
| expression $$(-2)^x=y$$ | evaluation | quality | log translation $$\log_{-2}y=x$$ |
| $-2^{-3}$ | $\frac{1}{-8}$ | – | $\log_{-2}(-\frac{1}{8})=-3$ |
| $-2^{-2}$ | $\frac{1}{4}$ | + | $\log_{-2}(\frac{1}{4})=-2$ |
| $-2^{-1}$ | $\frac{1}{-2}$ | – | $\log_{-2}(-\frac{1}{2})=-1$ |
| $-2^{0}$ | $1$ | + | $\log_{-2}(1)=0$ |
| $-2^{1}$ | $-2$ | – | $\log_{-2}(-2)=1$ |
| $-2^{2}$ | $4$ | + | $\log_{-2}(4)=2$ |
| $-2^{3}$ | $-8$ | – | $\log_{-2}(-8)=3$ |
| $-2^{4}$ | $16$ | + | $\log_{-2}(16)=4$ |
So, certainly there are some solutions to $log_{-2}x=y$. For example, as taken from the chart above, $\log_{-2}-8=3$ and $\log_{-2}-2=1$ and so on. However, $log_{-2}2$ has no solution, since there is no exponent such that $(-2)^x=-2$.
Of course, fractional exponents become a problem if we are only considering the Real Numbers. Consider
$$(-2)^{1/2}\longrightarrow\sqrt{-2}$$
As we previously learned, there is no real number that is the square root of a negative number. However, there are real solutions in the odd indexed roots; consider
$$\sqrt[3]{-8}=-2\longrightarrow\,(-2)(-2)(-2)=(4)(-2)=-8$$
Generally, we can take the roots of negative numbers when the index is odd, but not when the index is even.
So, there are integers on $x$ in $\log_{-2}x$ where this works and where this does not work. For example $x=1$ works for $\log_{-2}x$ but not for $x=2$ or $x=3$; however once we get to $x=4$.
Consequently, taking the log of any negative base is a fraught question, but one that is sometimes bears Real answers; however, these Real answers are sparse among the very many non-real answers.
4. Practice Problems with Solutions
Evaluate: $\log_{3}9\quad\log_{2}32\quad\log_{5}\frac{1}{125}\quad\log_{\frac{1}{2}}64\quad\log_{\frac{2}{3}}\frac{8}{27}$
$\log_{3}9$
First, it might make things easier if you re-write the problem in exponential form: $3^?=9$
We might now run through the various powers of 3 that we know and can easy calculate. Turns out that 2 is the power we will need since $3^2=9$.
Thus, we can say that $log_{3}9=2$
$\log_{2}32$
First, it might make things easier if you re-write the problem in exponential form: $2^?=32$
We might now run through the various powers of 2 that we know and can easy calculate:
$$2^1=2\quad2^2=(2)(2)=4\quad2^3=(2)(2)(2)=8\quad2^4=(2)(2)(2)(2)=16\quad2^5=(2)(2)(2)(2)(2)=32$$
Turns out that 5 is the power we will need since $2^5=32$.
Thus, we can say that $log_{2}32=5$
$\log_{5}\frac{1}{125}$
First, it might make things easier if you re-write the problem in exponential form: $5^?=\frac{1}{125}$
We first must make the logical leap that to transform an integer like 5 into a fraction via exponentiation, we must use negative exponents: i.e. $a^{-b}=\frac{1}{a^b}$ for some integers $a$ and $b$ where $b\gt0$.
We might now run through the various powers of 5 that we know and can easy calculate. Turns out that 3 is the power we will need since $5^3=125$.
Thus, since $5^{-3}=\frac{1}{5^3}=\frac{1}{125}$ , we can say that $log_{5}125=-3$
$\log_{\frac{1}{2}}64$
First, it might make things easier if you re-write the problem in exponential form:
$$\left(\frac{1}{2}\right)^?=64$$
As in the previous problem, we can recognize that the exponent must be negative; however, this time we use the negative exponent to transform a fraction into a integer: i.e. $\left(\frac{1}{x}\right)^{-1}=x\quad$.
So now that we know the exponent we need is negative, we should run through the various powers of 2 that we know and can easy calculate. We might recall from the first question that $2^5=32$, and so by merely multiplying $2^5$ by an additional $2$, we get $64$. Thus $2^6=64\quad$.
Thus, we can say that $\log_{\frac{1}{2}}64=6$
$\log_{\frac{2}{3}}\frac{8}{27}$
First, it might make things easier if you re-write the problem in exponential form:
$$\left(\frac{2}{3}\right)^?=\frac{8}{27}$$
Since $2^3=8$ and $3^3=27$, we can say that $\left(\frac{2}{3}\right)^3=\left(\frac{2^3}{3^3}\right)=\frac{8}{27}$.
This is true by the Power of a Quotient Rule (See “Five Laws of Exponents”). Thus,
$\log_{\frac{2}{3}}\frac{8}{27}=3$
Evaluate: $\log1000\quad\log\frac{1}{10}\quad\ln{e}\quad\ln{\sqrt[3]{e}}\quad\log_{\pi}1$
$\log1000$
Recall that when there is no base specified, we assume that this is the “common log,” which has a base of 10.
Since $10^3=1000$, we know that $\log1000=3$
$\log\frac{1}{10}$
Again, this is the common log with base 10. Since we must raise 10 to a power and return a fraction, we know the exponent must be negative.
Since $10^-1=\frac{1}{10^1}=\frac{1}{10}$, we know that $log\frac{1}{10}=-1$
$\ln{e}$
This is the “natural log,” which has a base of $e$, or Euler’s Number. Thus we are asking $\log_{e}e=?$
Translated into an exponential expression, we are asking $e^?=e$. Well we know that anything raise to the first power will return itself: i.e. $x^1=x$ for all $x$.
Since $e^1=e$, we know that $\ln{e}=1$
$\ln{\sqrt[3]{e}}$
The best way to approach this is to translate the rooting operation into an exponentiation operation via fractional exponents: $\sqrt[3]e=e^{\frac{1}{3}}$. Thus we can re-write our problem as
$$\ln{e^\frac{1}{3}}$$
Since all logs are merely asking what is the exponent I will need to apply to the base to return the given value, the exponent in the given value here is our answer. We might be able to see this better as an exponentiation operation.
$$\ln{e^\frac{1}{3}}\quad\longrightarrow\quad{e^?=e^{\frac{1}{3}}}$$
Since $e^{\frac{1}{3}}=e^{\frac{1}{3}}$, we know that $\ln{\sqrt[3]{e}}=\tfrac{1}{3}$
$\log_{\pi}1$
The only way to exponentiate a number (other than 1) and return 1 as a result, is to raise that number to the zero power: i.e. $x^0=1$ for all $x$ in $\mathbb{R}$.
Since $\pi$ is a Real Number, $\pi^0=1$. Thus, we know that $\log_{\pi}1=0$