How Logarithms and Exponents are Related

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From the course by Duke University

Data Science Math Skills

1265 ratings

Duke University

Data Science Math Skills

1265 ratings

Data science courses contain math—no avoiding that! This course is designed to teach learners the basic math you will need in order to be successful in almost any data science math course and was created for learners who have basic math skills but may not have taken algebra or pre-calculus. Data Science Math Skills introduces the core math that data science is built upon, with no extra complexity, introducing unfamiliar ideas and math symbols one-at-a-time. Learners who complete this course will master the vocabulary, notation, concepts, and algebra rules that all data scientists must know before moving on to more advanced material. Topics include: ~Set theory, including Venn diagrams ~Properties of the real number line ~Interval notation and algebra with inequalities ~Uses for summation and Sigma notation ~Math on the Cartesian (x,y) plane, slope and distance formulas ~Graphing and describing functions and their inverses on the x-y plane, ~The concept of instantaneous rate of change and tangent lines to a curve ~Exponents, logarithms, and the natural log function. ~Probability theory, including Bayes’ theorem. While this course is intended as a general introduction to the math skills needed for data science, it can be considered a prerequisite for learners interested in the course, "Mastering Data Analysis in Excel," which is part of the Excel to MySQL Data Science Specialization. Learners who master Data Science Math Skills will be fully prepared for success with the more advanced math concepts introduced in "Mastering Data Analysis in Excel." Good luck and we hope you enjoy the course!

From the lesson

Measuring Rates of Change

This module begins a very gentle introduction to the calculus concept of the derivative. The first lesson, "This is About the Derivative Stuff," will give basic definitions, work a few examples, and show you how to apply these concepts to the real-world problem of optimization. We then turn to exponents and logarithms, and explain the rules and notation for these math tools. Finally we learn about the rate of change of continuous growth, and the special constant known as “e” that captures this concept in a single number—near 2.718.

Using Integer Exponents7:48

Simplification Rules for Algebra using Exponents11:08

How Logarithms and Exponents are Related12:36

The Change of Base Formula4:23

The Rate of Growth of Continuous Processes11:09

Meet the Instructors

Daniel Egger
Executive in Residence and Director, Center for Quantitative Modeling
Pratt School of Engineering, Duke University
Paul Bendich
Assistant research professor of Mathematics; Associate Director for Curricular Engagement at the Information Initiative at Duke
Mathematics

Once you have a basic understanding of exponents,

it should not be too much of a leap to understand logarithms.

The reason why is that these two concepts are very closely related.

A logarithm is really the answer to the question raised to what power?

So for example, if you have 2 times 2 times 2 or 8, and

you want to know what power is being use, well, it's 2 to the 3rd.

In just the same way,

the logarithm to the base 2 of 8 is 3.

The logarithm to the base 2 of 16 would be 4.

The logarithm to the base 2 of 32 would be 5 and so on.

So, we'll be talking about logarithms the next couple videos.

And I hope that after you've worked some examples,

you'll see that just like exponents they follow some simple basic rules and

they're really not as intimidating as they might have seemed.

So, a formula for logarithms can have two different forms.

It can have the exponential form where we have

a number raised to the power x and this equals something.

So, to use the example that I just used,

you can have b=2, x= 3 and

2 to the 3=8, which is N.

Or, we can express the same idea in logarithmic form.

And here, we write log to the base b.

We use the letter b,

because we are referring to the base of a logarithm of the number N.

So log to the base 2(8)=X=3,

and that's the logarithmic

form of the same idea,

namely that 2 to the 3=8.

So 2 to the 4=16, log2(16)=4.

What is b?

b is the base.

So that's 2.

What is N?

N is 16.

And what is the exponent X?

That would be 4.

Okay, so we have the exponential form and the logarithmic form,

and they capture the same idea which is the relationship

between three numbers, a base, an exponent and the result.

So, just as we have the simple rule when raising any number to the power 0,

any number to the power 0, you may remember is equal to 1.

Similarly, the log to any base,

this number here is the base, of 1 is equal to 0.

Right, why?

Because 2 to the 0=1.

10 to the 0=1.

20 to the 0=1, when expressed in exponential form.

Just as with exponents, we have a few basic rules

that we will use to solve and simplify problems that contain logarithms.

Our three primary rules are the product rule, the quotient rule and

the power and root rule.

The product rule looks like this.

Log of a product

(XY).

Now when I write log without a base,

what this notation means is that this is true for any base.

So, this is a general formula and I don't need to specify the base as long as I

keep the base the same within any given problem.

So any log(XY)

= log X + log Y.

So, let's do a quick example.

We have logb(35).

This is equal to log b(7) + log b(5),

because 7 times 5 equals 35.

Very similarly, the quotient

rule says that if I have the log of

a fraction or a quotient, X over Y,

this is equal to log X- log Y.

So if I had log of 64 divided by 2,

this would be the same as log of 64 minus log of 2.

Power and Root Rule is the rule that we apply when we have

a number raised to a power, and we want to take its logarithm.

This is equal to nlogx.

So there are many situations when this makes it quite easy to solve an otherwise

hard problems.

So for example, we might have the log of the square root of some number and

this is simply going to be one-half the log of that number.

This rules works when N is positive or negative, an integer or a fraction.

So we know that 35=7 times five.

So this will be equal to

logb(7) + logb(5).

35 is also equal to 70/2.

So this would also be equal to

logb(70)- logb(2).

Log2(16/4).

Well, that should be obvious that,

that would be equal to log2(16)- log2(4).

Log2(16) is 2 raised to what power equals 16?

So, that would be 2 times 2 times 2 times 2, or 2 to the 4.

Log2(4) would be 2 raised to what power equals 4?

So that would be 2 times 2, or 2 to the 2, okay?

So log2(16/4) is going

to be equal to 4 minus 2 or 2.

Okay, let's look at a few more examples.

Log2(1,000), cube root of 1,000,

would simply be one-third log2(1,000)

Log10(7) to the fifth would

simply be 5 times log10(7).

Log of anything of x to the -1,

would be- logb of x.

So, those are our basic rules.

And if you apply them, you can simplify just about any logarithm.

So let's try combining several rules together at the same time, okay?

We're going to use our product and root rule, but

first we're going to use our product rule.

So we know that this is

equal to logb of x squared

+ logb of y to the -3,

which of course is

equal to 2logb of x-

3logb of y, okay?

Now, we apply our quotient rule.

So, logb of x squared divided by x to the minus

one-half is going to be equal to logb of x

squared minus logb of x to the minus one-half.

So, we have 2logb of x.

And now we have a minus, a minus which is a plus.

+ one-half logb of y, okay?.

Here's a handy trick just in general.

When we have log

a + log b = log

of a times b.

It's also going to b equal to log a- log of 1 over b.

So we simply put in a minus 1 and a minus 1.

And there are problems where that is very useful.

One of the sophisticated ways that we can use logarithms

is to treat both sides of a logarithmic equation as if we had an exponent.

So we have an equation, either a logarithmic or

an exponential equation of the form x=y, and

we treat both sides of that equation as if they were exponents of the same number.

Let me show you what I mean.

Okay, we have log2(39x/x-5)=4.

So, what we're going to do is we're going to

just take 2 to the log2 of all of this.

= 2 to the 4.

So, 2 to the log2 of anything simply equals that thing.

So we have 39x/x-5=16.

Then we can multiply both sides of this

equation by x-5, and we can simplify.

So we have 39x=16x.

80.

Now, we subtract 16x from both sides.

So, we have 23x = -80.

We have x = -80/23.

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