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.

Happy Valentines Day 2020

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Happy Valentines Day!
Or Happy St. Valentines Day, if you prefer.

I didn't know whether to make a Huey Lewis, Celine Dion, or Air Supply reference, so I just mentioned all three of them. That's the News, I'm sticking with it.




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Power Surge

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Nothing kills a ''quick'' joke like pixelation of older images. Sigh.
Originally, I was just going to substitute the other AnthroNumeric(tm) characters, but then I decided I should match the colors and faces. Ooooh, bad decision.

If you take numbers ending in 1 to 9 and raise them to the second power, they will end in 1, 4, 9, 6, 5, 6, 9, 4, 1.
If you raise them to the third power, they will end in 1, 8, 7, 4, 5, 6, 3, 2, 9.
If you raise them to the fourth power, they will end in 1, 6, 1, 6, 5, 6, 1, 6, 1.
And, finally, if you raise them to the fifth power, you return to 1, 2, 3, 4, 5, 6, 7, 8, 9, and the cycle will start over again.

The interesting things here:
In 1, 5, 6, and 0, if included, never change.
Row 2 is symmetrical (if you leave out 0). Also, no perfect square ends in 2, 3, 7 or 8.
Row 3 uses all nine numbers from 1 to 9.
Row 4 is also symmetrical and only contains 3 numbers. (That pesky 5!)





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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.

Power Station

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Afterward you can divide. Are you gonna do it?
As an element of the set of Real numbers, I know you qualify. Are you gonna do it?

Feel the Math.

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Grayscale. Graph skill. Grade scalar.
You make the pun.

Math and Religion

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Seriously, how could any mathematician not believe in "higher powers"?



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