# Exponentials & logarithms ### The inverse relationship of exponents and logarithms For $$m>0$$ and $$b>0, b\neq 1$$ we have the following relationship: $$\huge\fbox{\blueD{ b^{\purpleD{{ q}}}}=\redD m \quad \underline{\text{if and only if} }\quad \log\_{\blueD b }{\redD m}=\purpleD q}$$ ### Explanations $$log(n^{100}) => \Theta(logN) == 100log(n)$$ --> power becomes constant $$log\_5(n) => \Theta(log N) == \frac{log(N)}{log\_5}$$--> log of 5 becomes denominator $$log\_3(81) = x$$ to which power do I have raise log to the base of 3 to get to 81 = 4 $$log\_{100}(1) = x ==> 100^x = 1 ==> x=0$$ $$log\_b(a) = c <==> b^c=a$$ * b is the base * c is the exponent * a is called the argument When rewriting an exponential equation in log form or a log equation in exponential form, it is helpful to remember that the base of the logarithm is the same as the base of the exponent. $$log\_2(64) = 6 <==> 2^6 = 64$$ Let's start by setting that expression equal to x. $$log\_4(64) = x$$ Writing this as an exponential equation gives us the following: $$4^x = 64$$ 4 to what power is 64? Well, $$4^3 = 64$$and so $$log\_4(64) = 3.$$ Just by asking, "**4 to what power is 64**?" ### Restrictions on the variables $$log\_b(a)$$ is defined when the base b is positive—and not equal to 1—and the argument a is positive. These restrictions are a result of the connection between logarithms and exponents. | Restriction | Reasoning | | ----------- | --------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------------- | | $$b>0$$ | In an exponential function, the base b is always defined to be positive. | | $$a>0$$ | $$log\_b(a)=c$$ means that $$b^c=a$$. Because a positive number raised to any power is positive, meaning $$b^c>0$$, it follows that $$a>0$$. | | $$b\neq1$$ | Suppose, for a moment, that b *could* be 1. Now consider the equation $$log\_1(3)=x$$. The equivalent exponential form would be $$1^x=3$$. But this can never be true since 1 to any power is always 1. So, it follows that $$b\neq1$$. | ### The common and natural logarithm | Name | Base | Regular notation | Special notation | | ----------------- | ---- | ---------------- | ---------------- | | Common logarithm | 10 | $$log\_10(x)$$ | $$log(x)$$ | | Natural logarithm | e | $$log\_e(x)$$ | $$ln(x)$$ | $$e$$ is a mathematical constant. It is an irrational number that is approximately equal to 2.718. It appears in many contexts that involve limits, which you will likely learn about as you study calculus. For now, just treat $$e$$ as you would any other number. While the notation is different, the idea behind evaluating the logarithm is exactly the same! Here are two examples with solutions. **Example 1** Evaluate $$log(100)$$ **Solution 1** By definition, $$log(100) = log\_{10}(100)$$ 10 to what power is 100? $$10^2=100$$, so $$log(100) = 2$$ **Example 2** Evaluate $$ln(e^3)$$ **Solution 2** By definition, $$ln(e^3)=log\_e(e^3)$$. $$e$$to what power is $$e^3$$? $$e^3=e^3$$, so $$ln(e^3)=3$$. ### Exercise $$log\_8(2) = x\newline 8^x=2\newline x=\frac{1}{3}\newline 8^{\frac{1}{3}}=2$$ $$log\_8(\frac{1}{2})=x\newline 8^{-\frac{1}{3}}=\frac{1}{8^\frac{1}{3}}=\frac{1}{2}\newline x=-\frac{1}{3}$$ ### Evaluate log (adv) $$log\_{27}(\frac{1}{3})=x$$ 1. The equation $$log\_b(y)=x$$ and $$b^x=y$$mean exactly the same thing according to the definition of logarithms. Therefore, we can rewrite our question as an exponential equation. *Do remember that for the equations to be equivalent, we need* $$y$$*and* $$b$$ *to be positive numbers, and* $$b\neq1$$ So if $$log\_{27}(\frac{1}{3})=x$$, then $$27^x=\frac{1}{3}$$ 2. 27 to what power is $$\frac{1}{3}$$ 3. Notice that the cube root of 27 is 3 and that $$3^{-1}=\frac{1}{3}$$ Since $$a^{\frac{m}{n}}=(\sqrt\[n]a)^m$$, it follows that $$(\sqrt\[3]{27})^{-1} = 27^{-\frac{1}{3}}=\frac{1}{3}$$ So $$log\_{27}(\frac{1}{3})=-\frac{1}{3}$$ **Note**: If answering the above question was difficult, you can solve the equation $$27^x=\frac{1}{3}$$to get the answer. $$27^x=\frac{1}{3} \newline(3^3)^x=3^{-1} \newline3^{3x}=3^{-1} \newline3x=-1 \newline x=-\frac{1}{3}$$ Another test: $$log\_{\frac{1}{3}}(81)=x \newline {\frac{1}{3}}^x=81 \newline (3^{-1})^x=3^4 \newline 3^{-x}=3^4 \newline -x=4 \newline x=-4$$ ### Intro to logarithm properties | | | | --------------------- | -------------------------------------------------------------------------------------------------------- | | **The product rule** | $$\large\log\_b(MN)=\log\_b(M)+\log\_b(N)$$ | | **The quotient rule** | $$\large\log\_b\left(\frac{M}{N}\right)=\log\_b(M)-\log\_b(N)$$ | | **The power rule** | $$\large\log\_b(M^p)=p\log\_b(M)$$ | | **The base rule** | $$\large{\log\_\blueD{b}(\purpleC a)=\dfrac{\log\_\greenE{x}(\purpleC a)}{\log\_\greenE{x}(\blueD b)}}$$ | \ (These properties apply for any values of $$M$$, $$N$$, and $$b$$ for which each logarithm is defined, which is $$M$$, $$N>0$$ and $$0\ $$\begin{aligned}\log\_2({4\cdot 8})&=\log\_2(4)+\log\_2(8)\ \ \log\_2(32)&=\log\_2(4)+\log\_2(8)&&\small{\gray{\text{Since $4\cdot8=32$}}}\ \ 5&=2+3&&\small{\gray{\text{Evaluating the logs}}}\ \ 5&=5\end{aligned}$$ This is by no means a proof! Rather, it may convince us that the property is plausible and perhaps give us some insight as to why this is true. We can use the product rule to rewrite logarithmic expressions. #### Example 1: Expanding logarithms For our purposes, *expanding* a logarithm means writing it as the sum of two logarithms or more. Let's expand $$\log\_6(5y)$$. Notice that the two factors of the argument of the logarithm are $$\blueD 5$$ and $$\greenD y$$. We can directly apply the product rule to expand the log. > l$$og\_6​(5y)​=log\_6​(5⋅y)\newline \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =log\_6​(5)+log\_6​(y)​​ \Rightarrow \text{Product rule​}$$ #### Example 2: Condensing logarithms For our purposes, *compressing* a sum of two or more logarithms means writing it as a single logarithm. Let's condense $$\log\_3(10)+\log\_3(x)$$. Since the two logarithms have the same base (base-3), we can apply the product rule in the reverse direction: > $$log\_3​(10)+log\_3​(x)​=log\_3​(10⋅x)=log\_3​(10x)​$$ #### An important note When we compress logarithmic expressions using the product rule, the bases of all the logarithms in the expression **must** be the same. For example, we cannot use the product rule to simplify something like $$\log\_2(8)+\log\_3(y)$$. ### The quotient rule: $$\log\_b\left(\dfrac{M}{N}\right)=\log\_b(M)-\log\_b(N)$$ This property says that the log of a quotient is the difference of the logs of the dividend and the divisor. > If $$M=81$$, $$N=3$$ and $$b=3$$, then according to the property, $$\log\_3\left(\dfrac{81}{3}\right)=\log\_3(81)-\log\_3(3)$$. > > The work below shows that the property is indeed true in this case! > > > $$\begin{aligned} \\\log\_3\left({\dfrac{81}{3}}\right)&=\log\_3(81)-\log\_3(3)\ \ \log\_3(27)&=\log\_3(81)-\log\_3(3)&&\small{\gray{\text{Since $81\div 3=27$}}}\ \ 3&=4-1&&\small{\gray{\text{Evaluating the logs}}}\ \ 3&=3\end{aligned}$$ > > This is by no means a proof! Rather, it may convince us that the property is plausible and perhaps give us some insight as to why this is true. Now let's use the quotient rule to rewrite logarithmic expressions. #### Example 1: Expanding logarithms Let's expand $$\log\_7\left(\dfrac{a}{2}\right)$$, writing it as the difference of two logarithms by directly applying the quotient rule. > $$\begin{aligned}\log\_7\left(\dfrac{\purpleC a}{\goldD 2}\right)&=\log\_7(\purpleC a)-\log\_7(\goldD 2) &\small{\gray{\text{Quotient rule}}} \end{aligned}$$ #### Example 2: Condensing logarithms Let's condense $$\log\_4(x^3)-\log\_4(y)$$. Since the two logarithms have the same base (base-4), we can apply the quotient rule in the reverse direction: > $$\begin{aligned}\log\_4(\purpleC{x^3})-\log\_4(\goldD{y})&=\log\_4\left(\dfrac{\purpleC{x^3}}{\goldD{y}}\right)&&\small{\gray{\text{Quotient rule}}}\ \ \end{aligned}$$ #### An important note When we compress logarithmic expressions using the quotient rule, the bases of all logarithms in the expression **must** be the same. For example, we cannot use the quotient rule to simplify something like $$\log\_2(8)-\log\_3(y)$$. ### The power rule: $$\log\_b(M^p)=p\log\_b(M)$$ This property says that the log of a power is the exponent times the logarithm of the base of the power. If $$M=4$$, $$p=2$$, and $$b=4$$, then according to the property, $$\log\_4\left(4^2\right)=2\log\_4(4)$$. The work below shows that the property is indeed true in this case! > $$\begin{aligned} \\\log\_4\left({4^2}\right)&=2\log\_4(4)\ \ \log\_4(16)&=2\log\_4(4)&&\small{\gray{\text{Since $4^2=16$}}}\ \ 2&=2\cdot 1&&\small{\gray{\text{Evaluating the logs}}}\\\ \ 2&=2 \end{aligned}$$ This is by no means a proof! Rather, it may convince us that the property is plausible and perhaps give us some insight as to why this is true. Now let's use the power rule to rewrite log expressions. #### Example 1: Expanding logarithms For our purposes in this section, *expanding* a single logarithm means writing it as a multiple of another logarithm. Let's use the power rule to expand $$\log\_2\left(x^3\right)$$. > $$\begin{aligned}\log\_2\left(x^\maroonC3\right)&=\maroonC3\cdot \log\_2(x)&&\small{\gray{\text{Power rule}}}\ \ &=3\log\_2(x) \end{aligned}$$ #### Example 2: Condensing logarithms For our purposes in this section, *condensing* a multiple of a logarithm means writing it as a another single logarithm. Let's use the power rule to condense $$4\log\_5(2)$$, When we condense a logarithmic expression using the power rule, we make any multipliers into powers. > $$\begin{aligned}\maroonC4\log\_5(2)&=\log\_5\left(2^\maroonC 4\right)\~\~&&\small{\gray{\text{Power rule}}}\ \ &=\log\_5(16)\ \end{aligned}$$ ### Evaluating logarithms: change of base rule $$\large log\_ab =\frac{log\_xb}{log\_xa}$$ $$log\_5 100 = \frac{log\_{10} 100}{log\_{10} 5} = \frac{2}{log\_{10}5} \approx 2.861$$ Suppose we wanted to find the value of the expression $$\log\_2(50)$$. Since 50 is not a rational power of 2, it is difficult to evaluate this without a calculator. However, most calculators only directly calculate logarithms in base-$$10$$ and base-$$e$$. So in order to find the value of $$\log\_2(50)$$, we must *change the base* of the logarithm first. ### The change of base rule We can change the base of any logarithm by using the following rule: $$\large{\log\_\blueD{b}(\purpleC a)=\dfrac{\log\_\greenE{x}(\purpleC a)}{\log\_\greenE{x}(\blueD b)}}$$ Notes: * When using this property, you can choose to change the logarithm to *any* base $$\greenE x$$. * As always, the arguments of the logarithms must be positive and the bases of the logarithms must be positive and not equal to 1 in order for this property to hold! ### Justifying the change of base rule At this point, you might be thinking, "Great, but *why* does this rule work?" > $$\log\_b(a)=\dfrac{\log\_x(a)}{\log\_x(b)}\qquad\leftarrow\goldD{\text{Change of base rule}}$$ To examine this, let's return to the original expression $$\log\_2(50)$$. If we let $$\log\_2(50)=n$$ then it follows that $$2^n=50$$. Because the two values are equal, we can take the log *in any base* of both sides. Now we have: > $$\begin{aligned} 2^n &= 50 \\\ \log\_x(2^n) &= \log\_x(50)&&\small{\gray{\text{If $Y=Z$, then $\log\_x(Y)=\log\_x(Z)$}}} \\\ n\log\_x(2)&=\log\_x(50)&&\small{\gray{\text{Power Rule}}}\\\ n &= \dfrac{\log\_x(50)}{\log\_x(2)} &&\small{\gray{\text{Divide both sides by $\log\_x(2)$}}}\end{aligned}$$ Since $$n=\log\_2(50)$$, we have that $$\log\_2(50)=\dfrac{\log\_x(50)}{\log\_x(2)}$$ as desired! By the same logic, we can prove the change of base rule. Just change $$2$$ to $$b$$ and $$50$$ to $$a$$ and you have your proof! **Which expression is equivalent to** $$\log(6)\cdot \log\_6(a)$$? Notice that the argument of the first logarithm is the same as the base of the second logarithm. If we apply the change of base rule, we will be able to cancel common factors. Since the base of the first logarithm is $$10$$, let's change the base of the second logarithm to $$10$$ as well. > $$\begin{aligned}\log( \maroonD 6)\cdot\log\_{ \maroonD 6}( a) &=\purpleC{\log(6)}\cdot \dfrac{ {\log(a)}}{\purpleC{\log(6)}}\\\ &=\purpleC{\cancel{\log(6)}}\cdot \dfrac{ {\log(a)}}{\purpleC{\cancel{\log(6)}}}\\\ &=\log(a) \end{aligned}$$
### Quiz **Which of the following is equivalent to** $$\dfrac{\log\_9(m)}{\log(m)}$$**?** Notice that the argument of the logarithm in the numerator is the same as the argument of the logarithm in the denominator. > $$\dfrac{\log\_9(\blueD m)}{\log(\blueD m)}$$ Since the base of the logarithm in the denominator is $$10$$, let's change the base of $$\log\_9(m)$$ > $$\begin{aligned}\dfrac{\log\_9(m)}{\log(m)}&={\log\_9(m)}\div \log(m)\\\ &=\dfrac{\log(m)}{\log(9)}\div \log(m) \end{aligned}$$ Division is multiplication by the reciprocal, and so this becomes: > $$\begin{aligned}\phantom{\dfrac{\log(3)}{\log\_n(3)}}&=\dfrac{\greenD{\log(m)}}{\log(9)}\cdot \dfrac{1}{\greenD{\log(m)}} \\\ \end{aligned}$$ Let's cancel common factors: > $$\large=\frac{\cancel{log(m)}}{log(9)}⋅\frac{1}{\cancel{log(m)}}=\frac{1}{log(9)}$$ In conclusion, $$\dfrac{\log\_9(m)}{\log(m)}=\dfrac{1}{\log(9)}$$.