Logarithm (Log) Rules and Their Matching Exponents Rules

One of the problems that I had in high school math was trying to remember lists of rules without knowing where the rules came from. This could have been because the rules weren't explained, weren't explained enough for me, or were taught when I was absent and I had to learn it from a dry text book. (Note: I recall missing over a week of school with the flu one year, and as a result had problems with trigonometry identities for the longest time. They started to make sense in college during a Calculus class when I was staring at a Unit Circle regarding some other topic.)

If the rules for logarithms were explained in the context for the rules for exponents, as shown below, I might not have struggled with them as much, because whatever I didn't recall from memorization, I would've been able to work out. Again, it's possible that these rules were explained, but I wasn't in class and I had to get someone else's notes or read it from the book.

Exponent Rules	Logarithm Rules
(xa)(xb) = xa + b	log ((xa)(xb)) = log xa + log xb
xa / xb = xa - b	log (xa / xb) = log xa - log xb
(xa)b = xab	log (xa)b = b log xa
x0 = 1 (x =/= 0)	logb 1 = 0
x1 = x	logb b = 1 (b > 0)
xn =/= 0, when x =/= 0	logb 0 is undefined
xn > 0, when x > 0	logb x is undefined when x < 0

So the rules for expanding or condensing logarithms come from the corresponding rules for exponents. It's just that the notation may make it look more confusing than it actually is.

The Logarithm base switch rule says log_bc = 1/log_cb. The equivalent with exponents can be shown with this example:

2³ = 8 and 8^1/3 = 2

The exponents in those two case are reciprocals.

log₂8 = 3 and log₈2 = 1/3

So log₂8 = 1 / log₈2, because the logs are reciprocals.

The Logarithm base change rule says log_bx = log_cx / log_cb.

As I recall, the main reason to derive this rule was because, back in the dark ages when I was in high school, we used log tables, which were printed in the back of a text book, and based on 10 (and there was probably another page based of natural log tables based on e). If you could convert the base to 10, then you could use the table.

There would also be instances when you had terms in different bases that had to be converted before you work with them.

This is one example that shows how it works:

log₂8 = 3
log₄8 = 3/2
log₄2 = 1/2 log₄8 / log₄2 = (3/2) / (1/2) = 3

If you divide log₁₀8 / log₁₀2, you would also get 0.90308.../0.30102..., which is 3.000.

If you divide ln 8 / ln 2, you would also get 2.07944.../0.6931..., which is 3.000.

Edits made to formatting, and fixing the consistency of the variables. I hit "publish" a little too soon.

Math's Got Problems

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(C)Copyright 2021, C. Burke. "AnthroNumerics" is a trademark of Christopher J. Burke and (x, why?).

The two problems are left as exercises for the reader. It's a math blog after all.

That said, there are answers. And I even worked them out. So it's your turn.

I also write Fiction! Check out In A Flash 2020, by Christopher J. Burke for 20 great flash fiction stories, perfectly sized for your train rides. Available in softcover or ebook at Amazon. If you enjoy it, please consider leaving a rating or review on Amazon or on Good Reads. Thank you.

Come back often for more funny math and geeky comics.

Blog: Zeroth Power and Fractional Exponents

I'm starting a new, old math book (report to come), and I came across something interesting in the first chapter about numbers: another way to look at why N⁰ = 1.

The rules for exponents are simple:

N^a * N^b = N^{(a + b)}
and N^a / N^b = N^{(a - b)}

In the case of multiplying repeatedly by the same number:

N^a * N = N^{(a + 1)}

So, say, 2⁴ * 2 = 2^{4 + 1} = 2⁵
and 2⁵ * 2 = 2^{5 + 1} = 2⁶

Likewise, when dividing, repeatedly, by the same factor, the exponent would be decremented:

Thus, 2³ / 2 = 2^{3 - 1} = 2²
and 2² / 2 = 2^{2 - 1} = 2¹
and, finally, 2¹ / 2 = 2^{1 - 1} = 2⁰.

So what is 2⁰?
If we evaluate the other expressions, we have the following:
8 / 2 = 4; 4 / 2 = 2; 2 / 2 = 1. So 2⁰ = 1.

Note that we could continue the progression into negative exponents if we keep dividing. This will lead to fractions. Perhaps on another day...

Fractional exponents

What does it mean to have a fraction as an exponent?

Keep in mind that negative exponents have nothing to do with negative numbers. They create fractions. So fractional exponents won't create fractions.

Let's review one more rule about exponents:

(N^a)^b = N^(ab)

So (2³)⁴ = 2^(3*4) = 2¹² = 4096.

What if there is an exponent of 2/3 or 3/5? First, consider that 2/3 = (2)(1/3) and 3/5 = (3)(1/5). The 2 and 3 still mean the second and third powers, but what about the unit fractions of 1/3 or 1/5?

If I take the positive square root, which I'll abbreviate SQRT(), of N², I will get N, because N * N = N².
If I take the positive square root of N⁴, I will get N², because N² * N² = N⁴, etc.

So SQRT(N^a) will give (N^(1/2)a).
However, our rule tells us that (N^(1/2)a) = (N^a)^1/2.
So (N^a)^1/2 is another way to write SQRT(N^a), and N^1/2 is another way to write SQRT(N).

Fractions and Zero

Consider the following progression:

SQRT(16) = 4
SQRT( SQRT(16)) = 2
SQRT( SQRT( SQRT(16))) = 1.414...
SQRT( SQRT( SQRT( SQRT(16)))) = 1.189...
SQRT( SQRT( SQRT( SQRT( SQRT(16))))) = 1.090...

As you repeatedly take the square root, the answer will get closer and closer to 1.
If we rewrite that using exponents, we get the following:

16^(1/2) = 4
(16^(1/2))^(1/2) = 16^(1/4) = 2
((16^(1/2))^(1/2))^(1/2) = 16^(1/8) = 1.414...
(((16^(1/2))^(1/2))^(1/2))^(1/2) = 16^(1/16) = 1.189...
((((16^(1/2))^(1/2))^(1/2))^(1/2))^(1/2) = 16^(1/32) = 1.090...

As the denominator gets larger, the fraction gets smaller. As the denominator goes toward infinity, the fraction goes toward zero. And the value on the right side of the equal sign goes toward 1.

Algebra 2 Problems of the Day

The following problems were taken from the ALGEBRA II (Common Core) Regents Exam given on Friday, January 27, 2017.
Previous problems can be found here

Part 1

7. The expression

is equivalent to

(2) (see below)

You may do some of these steps in a different order. You might combine several of them.
I'm breaking it down as much as I can so more people can follow along.
Here is my process:

Resolve the division by subtracting the exponents: 2 - (1/3) = 1 2/3 or 5/3.
Next resolve the exponents: to find a power or a power, multiply the exponents, so (5/3)(-1/2) = (-5/6)
A negative exponent indicates the reciprocal: m^(-5/6) = (1 / m^5/6)
A fractional exponent indicates the nth root: see illustration above.

8. What is the inverse of the function y = log ₃ x?

(3) y = 3^x
Logs are the inverse of exponential functions.




Continue to the next problems

Happy New Years Eve 2015

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(C)Copyright 2014, C. Burke.

At least he came to give moral support. Two to the thirteenth won't be around for a while!

Two notes: First, 2^11 - 2^5 - 1 = 2015. That is to say, if all the powers of two, up to 10, were included, the sum would be 2047, which is one less that 2 to the 11th, which is 2048. Second, let's not forget the Y2048 bug! Or maybe the Y2(11) bug, just to be creative. Yes, it's true -- some really old computers, which will be really, really old 33 years from now, will have a problem handling the year as 2 to the 11th power. It should prove as Earth-shattering as Y2K did.

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UPDATE: The Making of a Webcomic

This comic started with an 11-digit binary number, 10 ones and 1 zero. I thought it would be funnier if I used the powers of 2 instead, so I needed 11 twos. And then I decided to include the next power as an extra gag. So I needed to draw 12 twos.

Rather than use one of the two twos I usually use and instead of typing the twos, I decided to try on of the paint programs on my tablet and doodled them. I was worried that they might be too snakelike -- and then I was worried that I'd doodled a row of ducks. (I have to keep this in mind if I ever need ducks again.) Then I made the smaller numeric exponents from 12 on down to 0. Colored it in and emailed it to me PC.

Putting it together I realized that I had too many exponents and not enough twos! Oops! I made a mistake. BUT I PICKED THE WRONG MISTAKE! The problem wasn't that I didn't have enough 2s (and quickly created an extra). The problem was that if the lowest exponent was zero, then the highest exponent I needed was 11.

And somehow though all the checking and proofreading -- including all that stuff above (which I have since corrected) -- none of this popped into my head. Of course, moving to 2^11 power would be a bigger problem than moving to 2^12. Some things are stored as 10 bits (I don't know why, but they were) but nothing would be stored as 11 bits (well, maybe -- programmers are strange).

Anyway, the correction has been made. HAPPY NEW YEAR!!

N-RN.2 (Real Number System) - More Rational Exponents

Continuing my post from Tuesday, on the Real Number system and rational exponents, let's move on to Standard N-RN.2, which reads Rewrite expressions involving radicals and rational exponents using the properties of exponents.

There's a lot to consider under this standard, so I'll continue with rational exponents, i.e., fractions. What if we wanted to evaluate an expression like this?

We need to recognize that the radical 5 is the same as 5^1/2, so

The rules for exponents say to multiply the 1/2 and the 4, giving us 5² or 25.

We can take this further. Suppose we had

The cube root is the same as 1/3 power. So

We can evaluate 6³ as 6 * 6 * 6 = 216.

One more example: How would we handle

The fourth root becomes the 1/4 power.

Now we can get a little fancy with and deal with the multiplication of two fractions:

One final note: The answer doesn't always have to be a rational number. You may exchange one rational power for another, one root for a different one. Consider:

Problems could contain any combination of roots and improper fractions, which may or may not have a simple rational answer. But keep the calculator handy just in case you need to know the sixth root of 117,649. Showing your work, of course.

N-RN.1 (Real Number System) - The Meaning of Rational Exponents

While I'm trying to update every day in August, I might as well start taking a closer look at the Common Core standards, which have now been in place for one year in high school Algebra 1 classes in New York. The first standard I find is N-RN.1. The first N stands for Number and Quantity. Unfortunately, the RN stands for Real Number System and not Registered Nurse because the latter would be helpful when you got sick of all this! I could have helped that telling that joke -- I chose to tell it any way, if only because I had to look up what the letters meant, particularly the first "N".

The entire standard reads as follows: Explain how the definition of the meaning of rational exponents follows from extending the properties of integer exponents to those values, allowing for a notation for radicals in terms of rational exponents.

It then has two components, one for Algebra 1, the other for Algebra 2. The Algebra 1 component is Evaluate integers raised to rational exponents (Algebra 1 - V.9). Don't ask me about the "V.9", I've done enough searching for this column.

I used to start teaching each year with Order of Operations, something that all the students should have seen before, and yet seemed to forget about. They know an acronym, such as PEMDAS, but don't know what it means. Oh, they know what the six letters mean, but they don't get the concept. And even when they can explain the concept, when push comes to shove and the pressures on (and they're taking a quiz), you find them calculating from left to right as if they hadn't learned anything. I had to change that when we started welcoming calculators into the lesson (even before we started requiring them). The calculators were down the work for them, so they didn't have to learn it, right? Wrong. I just adapted the problems. I started added more operations within fractions and adding exponents, forcing them to pay attention to what they put into their calculator. For instance, you need explicit parentheses to group things in a calculator because the numerator and denominator are implicitly grouped.

But Common Core changes that. They have to learn it much earlier, so they're ready for Algebra by the time they get to high school. Yeah, right. I'm still doing it. But wait, there's more.

After parentheses, come the exponents. Some students know the concept of exponents, and some just press buttons on the calculator. They know (or they'll learn!) that they are multiplying the factor some number of times. What they haven't seen before is a fraction as an exponent. What do you do when you see a fraction? Hide under the desk, usually. Wait for it to be over. I'm not exaggerating much.

First, I have to teach the concept that exponent 1/2 means take the square root, as stated in the gem. And then see how well they know they're square roots, either with or without a calculator. (I try to get them to learn up to 256, with some success.)

Second, I have to teach them that exponent 1/3 means take the cube root. This usually entails explaining what a cube root is, and where to find it on the calculator. Maybe reviewing what "cubing" means, and possibly through in Volume = length X width X height somewhere, just so I can spiral back to it when that comes around again. Depending on the results, I could try to conquer the concepts of fourth and fifth roots. Seriously, a practice webpage asked me (729)^(1/6).

At this point, I'm asking: are they getting it, or are they pushing buttons? I don't mind the button pushing if they understand the concept, because then they'll start recognizing patterns and will be smarter about their button pushing.

Okay, so the next step is the real doozy: explaining exponents of 2/3 or 3/5. The students have to deconstruct the fraction. That is, they need to know that 2/3 = (2)X(1/3), so they need both to square it and then take the cube root. And then I'll suggest that you take the cube root first, so that they're dealing with smaller numbers. Sometimes this makes sense to them.

Finally, there's the kicker: improper fractions. If fractions are Dr. Frankenstein, improper fractions are his monster. They're a whole new level of scary, and the first thing they want to do is turn them into mixed numbers. Or decimals. No! Wait! Stop!

Taking the (5/2) power of 16 is as simple as taking the fifth power of the square root of 16. Okay, read that again like you're a ninth grader.

No, it's not that difficult to do once the concept is learned, but it's something they haven't seen before, and they're learning it earlier. I can't remember exactly when the first time I was required to find a third or fourth root. I was probably using logs to do it. But then, I didn't have the calculators that they have today on their phones. They'll have the answers at their fingertips, but I'll make them show the work so I know they get the idea behind it.