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On this page
  • The inverse relationship of exponents and logarithms
  • Explanations
  • Restrictions on the variables
  • The common and natural logarithm
  • Exercise
  • Evaluate log (adv)
  • Intro to logarithm properties
  • The quotient rule:
  • The power rule:
  • Evaluating logarithms: change of base rule
  • The change of base rule
  • Justifying the change of base rule
  • Quiz
  1. math
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Exponentials & logarithms

PreviousextrasNextTrigonometry

Last updated 5 years ago

The inverse relationship of exponents and logarithms

For m>0m>0m>0 and b>0,bโ‰ 1b>0, b\neq 1b>0,b๎€ =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(n100)=>ฮ˜(logN)==100log(n) log(n^{100}) => \Theta(logN) == 100log(n)log(n100)=>ฮ˜(logN)==100log(n) --> power becomes constant

log5(n)=>ฮ˜(logN)==log(N)log5 log_5(n) => \Theta(log N) == \frac{log(N)}{log_5} log5โ€‹(n)=>ฮ˜(logN)==log5โ€‹log(N)โ€‹--> log of 5 becomes denominator

log3(81)=x log_3(81) = x log3โ€‹(81)=x to which power do I have raise log to the base of 3 to get to 81 = 4

log100(1)=x==>100x=1==>x=0 log_{100}(1) = x ==> 100^x = 1 ==> x=0 log100โ€‹(1)=x==>100x=1==>x=0

logb(a)=c<==>bc=a log_b(a) = c <==> b^c=a logbโ€‹(a)=c<==>bc=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.

log2(64)=6<==>26=64 log_2(64) = 6 <==> 2^6 = 64 log2โ€‹(64)=6<==>26=64

Let's start by setting that expression equal to x.

log4(64)=x log_4(64) = x log4โ€‹(64)=x

Writing this as an exponential equation gives us the following:

Just by asking, "4 to what power is 64?"

Restrictions on the variables

Restriction

Reasoning

In an exponential function, the base b is always defined to be positive.

The common and natural logarithm

Name

Base

Regular notation

Special notation

Common logarithm

10

Natural logarithm

e

While the notation is different, the idea behind evaluating the logarithm is exactly the same!

Here are two examples with solutions.

Example 1

Solution 1

10 to what power is 100?

Example 2

Solution 2

Exercise

Evaluate log (adv)

Another test:

Intro to logarithm properties

The product rule

The quotient rule

The power rule

The base rule

What you should be familiar with before taking this lesson

What you will learn in this lesson

Logarithms, like exponents, have many helpful properties that can be used to simplify logarithmic expressions and solve logarithmic equations. This article explores three of those properties.

Let's take a look at each property individually.

This property says that the logarithm of a product is the sum of the logs of its factors.

The work below shows that the property is indeed true in this case!

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.

Example 2: Condensing logarithms

For our purposes, compressing a sum of two or more logarithms means writing it as a single logarithm.

Since the two logarithms have the same base (base-3), we can apply the product rule in the reverse direction:

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.

This property says that the log of a quotient is the difference of the logs of the dividend and the divisor.

The work below shows that the property is indeed true in this case!

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

Example 2: Condensing logarithms

Since the two logarithms have the same base (base-4), we can apply the quotient rule in the reverse direction:

An important note

When we compress logarithmic expressions using the quotient rule, the bases of all logarithms in the expression must be the same.

This property says that the log of a power is the exponent times the logarithm of the base of the power.

The work below shows that the property is indeed true in this case!

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.

\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.

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

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:

  • 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?"

Because the two values are equal, we can take the log in any base of both sides. Now we have:

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.

Quiz

Notice that the argument of the logarithm in the numerator is the same as the argument of the logarithm in the denominator.

Division is multiplication by the reciprocal, and so this becomes:

Let's cancel common factors:

4x=64 4^x = 64 4x=64

4 to what power is 64? Well, 43=64 4^3 = 64 43=64and so log4(64)=3. log_4(64) = 3. log4โ€‹(64)=3.

logb(a) log_b(a) logbโ€‹(a) is defined when the base b is positiveโ€”and not equal to 1โ€”and the argument a is positive. These restrictions are a result of the connection between logarithms and exponents.

means that . Because a positive number raised to any power is positive, meaning , it follows that .

Suppose, for a moment, that b could be 1. Now consider the equation . The equivalent exponential form would be . But this can never be true since 1 to any power is always 1. So, it follows that .

eee 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 eee as you would any other number.

Evaluate log(100)log(100)log(100)

By definition, log(100)=log10(100)log(100) = log_{10}(100)log(100)=log10โ€‹(100)

102=100 10^2=100102=100, so log(100)=2log(100) = 2log(100)=2

Evaluate ln(e3)ln(e^3)ln(e3)

By definition, ln(e3)=loge(e3)ln(e^3)=log_e(e^3)ln(e3)=logeโ€‹(e3).

eeeto what power is e3e^3e3?

e3=e3 e^3=e^3e3=e3, so ln(e3)=3ln(e^3)=3ln(e3)=3.

log8(2)=x8x=2x=13813=2log_8(2) = x\newline 8^x=2\newline x=\frac{1}{3}\newline 8^{\frac{1}{3}}=2log8โ€‹(2)=x8x=2x=31โ€‹831โ€‹=2

log8(12)=x8โˆ’13=1813=12x=โˆ’13log_8(\frac{1}{2})=x\newline 8^{-\frac{1}{3}}=\frac{1}{8^\frac{1}{3}}=\frac{1}{2}\newline x=-\frac{1}{3}log8โ€‹(21โ€‹)=x8โˆ’31โ€‹=831โ€‹1โ€‹=21โ€‹x=โˆ’31โ€‹

log27(13)=x log_{27}(\frac{1}{3})=xlog27โ€‹(31โ€‹)=x

The equation logb(y)=xlog_b(y)=xlogbโ€‹(y)=x and bx=yb^x=ybx=ymean 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 yyyand bbb to be positive numbers, and bโ‰ 1b\neq1b๎€ =1

So if log27(13)=xlog_{27}(\frac{1}{3})=xlog27โ€‹(31โ€‹)=x, then 27x=1327^x=\frac{1}{3}27x=31โ€‹

27 to what power is 13\frac{1}{3}31โ€‹

Notice that the cube root of 27 is 3 and that 3โˆ’1=133^{-1}=\frac{1}{3}3โˆ’1=31โ€‹

Since amn=(an)ma^{\frac{m}{n}}=(\sqrt[n]a)^manmโ€‹=(naโ€‹)m, it follows that (273)โˆ’1=27โˆ’13=13(\sqrt[3]{27})^{-1} = 27^{-\frac{1}{3}}=\frac{1}{3}(327โ€‹)โˆ’1=27โˆ’31โ€‹=31โ€‹

So log27(13)=โˆ’13log_{27}(\frac{1}{3})=-\frac{1}{3}log27โ€‹(31โ€‹)=โˆ’31โ€‹

Note: If answering the above question was difficult, you can solve the equation 27x=1327^x=\frac{1}{3}27x=31โ€‹to get the answer.

27x=13(33)x=3โˆ’133x=3โˆ’13x=โˆ’1x=โˆ’1327^x=\frac{1}{3} \newline(3^3)^x=3^{-1} \newline3^{3x}=3^{-1} \newline3x=-1 \newline x=-\frac{1}{3}27x=31โ€‹(33)x=3โˆ’133x=3โˆ’13x=โˆ’1x=โˆ’31โ€‹

log13(81)=x13x=81(3โˆ’1)x=343โˆ’x=34โˆ’x=4x=โˆ’4log_{\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=-4log31โ€‹โ€‹(81)=x31โ€‹x=81(3โˆ’1)x=343โˆ’x=34โˆ’x=4x=โˆ’4

(These properties apply for any values of MMM, NNN, and bbb for which each logarithm is defined, which is MMM, N>0N>0N>0 and 0<bโ‰ 10<b\neq10<b๎€ =1.)

You should know what logarithms are. If you don't, please check out our .

The product rule: logโกb(MN)=logโกb(M)+logโกb(N)\log_b(MN)=\log_b(M)+\log_b(N)logbโ€‹(MN)=logbโ€‹(M)+logbโ€‹(N)

If M=4M=4M=4, N=8N=8N=8 and b=2b=2b=2, then according to the property, logโก2(4โ‹…8)=logโก2(4)+logโก2(8)\log_2(4\cdot 8)=\log_2(4)+\log_2(8)log2โ€‹(4โ‹…8)=log2โ€‹(4)+log2โ€‹(8).

logโก2(4โ‹…8)=logโก2(4)+logโก2(8)logโก2(32)=logโก2(4)+logโก2(8)Sinceย 4โ‹…8=325=2+3Evaluatingย theย logs5=5\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}log2โ€‹(4โ‹…8)log2โ€‹(32)55โ€‹=log2โ€‹(4)+log2โ€‹(8)=log2โ€‹(4)+log2โ€‹(8)=2+3=5โ€‹โ€‹Sinceย 4โ‹…8=32Evaluatingย theย logsโ€‹

Let's expand logโก6(5y)\log_6(5y)log6โ€‹(5y).

Notice that the two factors of the argument of the logarithm are 5\blueD 55 and y\greenD yy. We can directly apply the product rule to expand the log.

log6โ€‹(5y)โ€‹=log6โ€‹(5โ‹…y)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย =log6โ€‹(5)+log6โ€‹(y)โ€‹โ€‹โ‡’Productย ruleโ€‹og_6โ€‹(5y)โ€‹=log_6โ€‹(5โ‹…y)\newline \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ =log_6โ€‹(5)+log_6โ€‹(y)โ€‹โ€‹ \Rightarrow \text{Product ruleโ€‹}og6โ€‹โ€‹(5y)โ€‹=log6โ€‹โ€‹(5โ‹…y)ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย =log6โ€‹โ€‹(5)+log6โ€‹โ€‹(y)โ€‹โ€‹โ‡’Productย ruleโ€‹

Let's condense logโก3(10)+logโก3(x)\log_3(10)+\log_3(x)log3โ€‹(10)+log3โ€‹(x).

log3โ€‹(10)+log3โ€‹(x)โ€‹=log3โ€‹(10โ‹…x)=log3โ€‹(10x)โ€‹log_3โ€‹(10)+log_3โ€‹(x)โ€‹=log_3โ€‹(10โ‹…x)=log_3โ€‹(10x)โ€‹log3โ€‹โ€‹(10)+log3โ€‹โ€‹(x)โ€‹=log3โ€‹โ€‹(10โ‹…x)=log3โ€‹โ€‹(10x)โ€‹

For example, we cannot use the product rule to simplify something like logโก2(8)+logโก3(y)\log_2(8)+\log_3(y)log2โ€‹(8)+log3โ€‹(y).

The quotient rule: logโกb(MN)=logโกb(M)โˆ’logโกb(N)\log_b\left(\dfrac{M}{N}\right)=\log_b(M)-\log_b(N)logbโ€‹(NMโ€‹)=logbโ€‹(M)โˆ’logbโ€‹(N)

If M=81M=81M=81, N=3N=3N=3 and b=3b=3b=3, then according to the property, logโก3(813)=logโก3(81)โˆ’logโก3(3)\log_3\left(\dfrac{81}{3}\right)=\log_3(81)-\log_3(3)log3โ€‹(381โ€‹)=log3โ€‹(81)โˆ’log3โ€‹(3).

logโก3(813)=logโก3(81)โˆ’logโก3(3)logโก3(27)=logโก3(81)โˆ’logโก3(3)Sinceย 81รท3=273=4โˆ’1Evaluatingย theย logs3=3\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}log3โ€‹(381โ€‹)log3โ€‹(27)33โ€‹=log3โ€‹(81)โˆ’log3โ€‹(3)=log3โ€‹(81)โˆ’log3โ€‹(3)=4โˆ’1=3โ€‹โ€‹Sinceย 81รท3=27Evaluatingย theย logsโ€‹

Let's expand logโก7(a2)\log_7\left(\dfrac{a}{2}\right)log7โ€‹(2aโ€‹), writing it as the difference of two logarithms by directly applying the quotient rule.

logโก7(a2)=logโก7(a)โˆ’logโก7(2)Quotientย 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}log7โ€‹(2aโ€‹)โ€‹=log7โ€‹(a)โˆ’log7โ€‹(2)โ€‹Quotientย ruleโ€‹

Let's condense logโก4(x3)โˆ’logโก4(y)\log_4(x^3)-\log_4(y)log4โ€‹(x3)โˆ’log4โ€‹(y).

logโก4(x3)โˆ’logโก4(y)=logโก4(x3y)Quotientย rule\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}log4โ€‹(x3)โˆ’log4โ€‹(y)โ€‹=log4โ€‹(yx3โ€‹)โ€‹โ€‹Quotientย ruleโ€‹

For example, we cannot use the quotient rule to simplify something like logโก2(8)โˆ’logโก3(y)\log_2(8)-\log_3(y)log2โ€‹(8)โˆ’log3โ€‹(y).

The power rule: logโกb(Mp)=plogโกb(M)\log_b(M^p)=p\log_b(M)logbโ€‹(Mp)=plogbโ€‹(M)

If M=4M=4M=4, p=2p=2p=2, and b=4b=4b=4, then according to the property, logโก4(42)=2logโก4(4)\log_4\left(4^2\right)=2\log_4(4)log4โ€‹(42)=2log4โ€‹(4).

logโก4(42)=2logโก4(4)logโก4(16)=2logโก4(4)Sinceย 42=162=2โ‹…1Evaluatingย theย logs2=2\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}log4โ€‹(42)log4โ€‹(16)22โ€‹=2log4โ€‹(4)=2log4โ€‹(4)=2โ‹…1=2โ€‹โ€‹Sinceย 42=16Evaluatingย theย logsโ€‹

Let's use the power rule to expand logโก2(x3)\log_2\left(x^3\right)log2โ€‹(x3).

Let's use the power rule to condense 4logโก5(2)4\log_5(2)4log5โ€‹(2),

logab=logxblogxa\large log_ab =\frac{log_xb}{log_xa}logaโ€‹b=logxโ€‹alogxโ€‹bโ€‹

log5100=log10100log105=2log105โ‰ˆ2.861log_5 100 = \frac{log_{10} 100}{log_{10} 5} = \frac{2}{log_{10}5} \approx 2.861log5โ€‹100=log10โ€‹5log10โ€‹100โ€‹=log10โ€‹52โ€‹โ‰ˆ2.861

Suppose we wanted to find the value of the expression logโก2(50)\log_2(50)log2โ€‹(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-101010 and base-eee. So in order to find the value of logโก2(50)\log_2(50)log2โ€‹(50), we must change the base of the logarithm first.

When using this property, you can choose to change the logarithm to any base x\greenE xx.

logโกb(a)=logโกx(a)logโกx(b)โ†Changeย ofย baseย rule\log_b(a)=\dfrac{\log_x(a)}{\log_x(b)}\qquad\leftarrow\goldD{\text{Change of base rule}}logbโ€‹(a)=logxโ€‹(b)logxโ€‹(a)โ€‹โ†Changeย ofย baseย rule

To examine this, let's return to the original expression logโก2(50)\log_2(50)log2โ€‹(50). If we let logโก2(50)=n\log_2(50)=nlog2โ€‹(50)=n then it follows that 2n=502^n=502n=50.

2n=50logโกx(2n)=logโกx(50)Ifย Y=Z,ย thenย logโกx(Y)=logโกx(Z)nlogโกx(2)=logโกx(50)Powerย Rulen=logโกx(50)logโกx(2)Divideย bothย sidesย byย logโกx(2)\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}2nlogxโ€‹(2n)nlogxโ€‹(2)nโ€‹=50=logxโ€‹(50)=logxโ€‹(50)=logxโ€‹(2)logxโ€‹(50)โ€‹โ€‹โ€‹Ifย Y=Z,ย thenย logxโ€‹(Y)=logxโ€‹(Z)Powerย RuleDivideย bothย sidesย byย logxโ€‹(2)โ€‹

Since n=logโก2(50)n=\log_2(50)n=log2โ€‹(50), we have that logโก2(50)=logโกx(50)logโกx(2)\log_2(50)=\dfrac{\log_x(50)}{\log_x(2)}log2โ€‹(50)=logxโ€‹(2)logxโ€‹(50)โ€‹ as desired!

By the same logic, we can prove the change of base rule. Just change 222 to bbb and 505050 to aaa and you have your proof!

Which expression is equivalent to logโก(6)โ‹…logโก6(a)\log(6)\cdot \log_6(a)log(6)โ‹…log6โ€‹(a)?

Since the base of the first logarithm is 101010, let's change the base of the second logarithm to 101010 as well.

logโก(6)โ‹…logโก6(a)=logโก(6)โ‹…logโก(a)logโก(6)=logโก(6)โ‹…logโก(a)logโก(6)=logโก(a)\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}log(6)โ‹…log6โ€‹(a)โ€‹=log(6)โ‹…log(6)log(a)โ€‹=log(6)โ€‹โ‹…log(6)โ€‹log(a)โ€‹=log(a)โ€‹

Which of the following is equivalent to logโก9(m)logโก(m)\dfrac{\log_9(m)}{\log(m)}log(m)log9โ€‹(m)โ€‹?

logโก9(m)logโก(m)\dfrac{\log_9(\blueD m)}{\log(\blueD m)}log(m)log9โ€‹(m)โ€‹

Since the base of the logarithm in the denominator is 101010, let's change the base of logโก9(m)\log_9(m)log9โ€‹(m)

logโก9(m)logโก(m)=logโก9(m)รทlogโก(m)=logโก(m)logโก(9)รทlogโก(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}log(m)log9โ€‹(m)โ€‹โ€‹=log9โ€‹(m)รทlog(m)=log(9)log(m)โ€‹รทlog(m)โ€‹

logโก(3)logโกn(3)=logโก(m)logโก(9)โ‹…1logโก(m)\begin{aligned}\phantom{\dfrac{\log(3)}{\log_n(3)}}&=\dfrac{\greenD{\log(m)}}{\log(9)}\cdot \dfrac{1}{\greenD{\log(m)}} \\\\ \end{aligned}lognโ€‹(3)log(3)โ€‹โ€‹=log(9)log(m)โ€‹โ‹…log(m)1โ€‹โ€‹

=log(m)log(9)โ‹…1log(m)=1log(9)\large=\frac{\cancel{log(m)}}{log(9)}โ‹…\frac{1}{\cancel{log(m)}}=\frac{1}{log(9)} =log(9)log(m)โ€‹โ€‹โ‹…log(m)โ€‹1โ€‹=log(9)1โ€‹

In conclusion, logโก9(m)logโก(m)=1logโก(9)\dfrac{\log_9(m)}{\log(m)}=\dfrac{1}{\log(9)}log(m)log9โ€‹(m)โ€‹=log(9)1โ€‹.

๐Ÿชถ
b>0b>0 b>0
a>0a>0a>0
logb(a)=clog_b(a)=clogbโ€‹(a)=c
bc=ab^c=abc=a
bc>0b^c>0bc>0
a>0a>0a>0
bโ‰ 1b\neq1b๎€ =1
log1(3)=xlog_1(3)=xlog1โ€‹(3)=x
1x=31^x=31x=3
bโ‰ 1 b\neq1b๎€ =1
log10(x)log_10(x)log1โ€‹0(x)
log(x)log(x)log(x)
loge(x)log_e(x)logeโ€‹(x)
ln(x)ln(x)ln(x)
logโกb(MN)=logโกb(M)+logโกb(N)\large\log_b(MN)=\log_b(M)+\log_b(N)logbโ€‹(MN)=logbโ€‹(M)+logbโ€‹(N)
logโกb(MN)=logโกb(M)โˆ’logโกb(N)\large\log_b\left(\frac{M}{N}\right)=\log_b(M)-\log_b(N)logbโ€‹(NMโ€‹)=logbโ€‹(M)โˆ’logbโ€‹(N)
logโกb(Mp)=plogโกb(M)\large\log_b(M^p)=p\log_b(M)logbโ€‹(Mp)=plogbโ€‹(M)
intro to logarithms
\large{\log_\blueD{b}(\purpleC a)=\dfrac{\log_\greenE{x}(\purpleC a)}{\log_\greenE{x}(\blueD b)}}