Math 102

[Getting A Life]

Yep, that's right, that's the paper, entry level math and I'm struggling with logs. I've never seen them before and now after 2 lectures I'm meant to have a handle on them. Summer school is a mad pace I can't let these slow me up too much.

 

I'm not here to cheat. here's something I solved, is there an easier way, this hurt my head. Is it just practise I need, why are these darn things so alien?

 

My attempt. Find r.

 

30®^6 = 300

 

r^6 = 10

 

6 log r = log 10

 

log r = log 10/6 = 1/6

 

log r = 1/6 -> 6^-1 = 1/6

 

10^(1/6) = r

 

r = 1.467799268

 

Very hard for me to bridge the gap and find meaning to log r = (1/6) when I finally clicked I'd found that statement in base 10 and could rearrange it so. Any way to see these things, or do I just need to keep remembering lists of powers? I'd have been screwed if I met this question in a test, it literally took me hours.

 

We all gotta start somewheres...

[Don Blazys]

To: Getting A Life

 

Quoting Getting A Life:

My attempt. Find r.

 

30®^6 = 300

 

r^6 = 10

 

Right then and there,

you could have simply taken the 6th root of both sides to get:

 

r=10^(1/6)=1.4677992676221...

 

Solving that particular problem did not require logarithms.

Just a good grasp of the basics and fundamentals of algebra,

which is the foundation upon which the rest of mathematics is built.

 

Make sure your foundation is solid.

In your spare time, read books on algebra... just for the fun of it.

 

Quoting Getting A Life:

Is it just practise I need, why are these darn things so alien?... this hurt my head

 

Logarithms are perhaps the most powerfull and underrated tool in all of mathematics.

 

They can bend your mind even after you do become an expert in their use.

 

Look at it this way...

 

If math was easy, then it wouldn't be any fun.

 

Would you play a video game that you could beat with your eyes closed?

 

Of course not! Such a game would be exeedingly boring,

 

and you would not waste your time with it.

 

Enjoy, relish and savor the challenge.

 

Don.

[Getting A Life]

Thank you Don.

 

I was not aware I could take a 6th root.

 

As for algebra, I really enjoy it but it does not come easily. I first met algebra in August last year in an adult bridging course. Damn was that a learning curve! I use a refill pad a week practising algebra but I still meet certain problems (like the above) and for some reason I'm blank.

 

In looking at some of tricks logs can perform I'm keen to get them straight in my head, but a new situation still throws my newly trained algebraic grey matter into a tailspin.

 

Logs, exponents, roots, my latest algebraic bugbears. I shall persevere as you are right - it's fun.

[CraigD]

Welcome to hypography, Getting A Life! :)

 

You might do well to try to fathom the “heart and soul / essence” or what a logarithm is. It’s almost certainly explained in you textbook, but teachers and students tend to skim past the essential definitions of things in their haste to begin doing problems with them. If you grasp the essentials, though, doing problems is much easier. You’ll have a feel for using them.

 

A logarithm is a numeric function, meaning if takes a number, and returns another number.

 

Its essential definition, IMO, is

 

[math]b^{\log_b(x)} = x[/math]

 

In words, this reads “b raised to the power of the log base b of x is x”.

 

[imath]b[/imath] is the base of the logarithm. There are an infinite number of logarithm functions, since they can have any base greater than 1. The most popular bases, however, are 10, (the log function base 10 is called the common logarithm, and is usually written “log”), [imath]e[/imath], a transcendental number that can be written with a finite number of digits, approximately 2.7182818 (the natual log, usually written “ln”), and 2 (the “binary log”, which has no well-known special name, so’s usually written just [imath]\log_2[/imath]).

 

If you have a log calculator of any base, it’s easy to get the log in any other base, because

 

[math]\log_a(x) = \frac{\log_b(x)}{\log_b(a)}[/math].

 

This is good, because the easiest log to calculate is the natural log. An easy (though not the most computationally efficient way) to approximate (except for special cases, you can’t calculate an exact natural log, because they’re transcendental) a natural log is by calculating as much of the following series:

 

[math]ln(x) = \frac{(x-1)^1}{1} -\frac{(x-1)^2}{1\cdot2} +\frac{(x-1)^3}{1\cdot2\cdot3} -\frac{(x-1)^4}{1\cdot2\cdot3\cdot4} + \dots[/math]

 

[imath]x[/imath] is the argument of the log function.

 

Though it’s not usually taught in introductory math courses, I think it’s a lot of fun to approximate logarithms using simple arithmetic, and encourage students to try it a few times to calculate some logs.

 

Once you’ve gotten a good mental lock on what a logarithm essentially is – something to raise a base to to get a number, how it works is pretty intuitive. Multiplying two numbers is can be done by adding their log, because:

 

[math]b^a \cdot b^c = b^{a+c}[/math]

 

Raising a number to a power (exponentiating) it can be done by multiplying its log, because:

 

[math]\left( b^a \right) ^c = b^{a \cdot c}[/math]

 

And so on.

 

Above all else, as with any math, try playing with this stuff, beyond just doing homework and prepping for tests. Nearly all people who know math well play with it a lot.

 

A final note about roots: they’re just number raised to fractional exponents. So [imath]\sqrt{a} = a^{\frac12}[/imath], [imath]\sqrt[3]{a} = a^{\frac13}[/imath], [imath]\sqrt[6]{a} = a^{\frac16}[/imath], etc.

[Getting A Life]

Wow, thanks Craig. Would you mind if I print this and show a couple of classmates?

 

I was messing with rational functions and derivatives last night and came up with this

 

A/B^c derived becomes A(-c)/B^C+1. Thing is it works half the time the other half a negative sign is in the wrong place. Any idea how to tidy this up? It might be ok it was very late I'll investigate it some more today.

 

Again, thank you I'm very grateful for the realistic advice and help you folks have given.

 

I never dreamed I'd enjoy math.

[CraigD]

Wow, thanks Craig. Would you mind if I print this and show a couple of classmates?

I’m flattered, Getting. :) Feel free to share this any way you like – this is the internet, after all, and information wants to be free!

A/B^c derived becomes A(-c)/B^C+1. Thing is it works half the time the other half a negative sign is in the wrong place.

This looks right to me, though it needs some parenthesis (or neater still, marking up with LaTeX:Math – click on any formula at hypography marked up this way, and it will show you the LaTeX used), and could use some rearranging and, to make it more legible. I read it as this:

[math]f(B) = \frac{A}{B^C} = A B^{-C}[/math]

[math]f '(B) = -CA B^{-C-1} = \frac{-AC}{B^{C+1}}[/math]

 

Expressions like B^C+1 are ambiguous, because they might mean [imath]1 + B^C[/imath] or [imath]B^{C+1}[/imath]. Most computer languages evaluate this as [imath]1+B^C[/imath]. To avoid confusion, use parenthesis, ie: B^(C+1).

 

It’s common to write derivatives like [imath] f '(B)[/imath] as [imath] \frac{df (B)}{dB}[/imath]. Either form is recognized by people who know calculus, and which one’s best depends IMHO on how you’re using it.

I never dreamed I'd enjoy math.

Math is some of the deepest, most fun stuff in the world. Lots of folk who love it madly. :)

[Getting A Life]

Again most grateful. We're getting two new subjects a day now, no problem with product or quotient rules but the chain rule is giving me grief.

 

The 1st example, after the working states

 

3/2x^2 x 4x = 6/x.

 

How did they get this. I can enter the values etc and get 3/2x^2 x 4x, but the algebra seems to take a leap to = 6/x; that I don't fathom. Without understanding this I fear I'll keep getting stuck at half an anwer ie: not enough working. Is there a name for what I am trying to fathom, I have a few books on math I can look it up.

 

ps: 90% on the midterm test, could improve, next time I won't leave early. Was a bit sick but poor excuse I had 20 minutes to check answers and didn't.

 

I'm also interested in Mayan math. I can multiply it out in base 10 and 20 now using both numeral systems. I'd love to know who has a handle on how they worked out division etc. Had very little time for it so far unfortunately, too much math 102.

[JMJones0424]

If I understand you correctly-

 

[math]

\frac {3}{2x^2} \cdot 4x

[/math]

 

[math]

\frac {3 \cdot 4 \cdot x}{2 \cdot x \cdot x}

[/math]

 

cancel out 4/2 and x/x. then

 

[math]

\frac {3 \cdot 2}{x} = \frac {6}{x}

[/math]

[Getting A Life]

Thanks JM. Methinks I'm getting a bit tired not to have spotted that. I'm taking the (1/2 an) evening off! Time for a movie and junk food.

[Getting A Life]

Yes, I needed the rest. There are lots of wee 'leaps' in the algebra, showing my lack of basics having skipped school early. I'm sitting with year 12 and 13 (with calculus) textbooks on my desk though and refusing to let anything slide.

 

Another brief question (or two):

 

If I derive 0.5x^2 does it become x or 1.

 

And... What is the order for sorting out a derivative. Do I use chain rule first then quotient/product rule, or vice versa.

 

Starting to get most of the chain rule problems now...

[Getting A Life]

More questions...

 

Where do I start to derive y= ln(x^5/5)

 

I'm having trouble with the x^5/5 term. I need a derivative. Ah just answered my own question, I think...

 

Do I use chain rule, and use quotient rule to get derivative of x^5/5 term, then solve?

 

Please don't solve that one for me, it's an assignment question. Just a hint if I'm moving in the right direction would be appreciated.

 

Also, if you have a term eg f (x) = 5 ; What value would I enter as f ' (x) ? 0?

 

Two weeks to go... got 2 extra marks awarded in test, marker was wrong - 94%. Thanks for the help! Exams gonna be tough though, lots of new material to learn yet and little time to do it in...

 

Nothing like a good challenge, the old gray matter is partially cobwebs but I'm soldiering on.

[Ben]

Yes, use the chain rule. Also use the following: if [math]y = \ln x[/math] then [math] x = ?[/math]

 

Then you will find that [math]\frac{d}{dy}?[/math] has an unusual property, which you should know, from which your answer follows instantly.

 

Please use the Tex capability provided here - it's not hard.

 

PS You ask a lot of questions!!

[Getting A Life]

Just a heads up on that paper.

 

THANK YOU VERY MUCH kind people, for helping me out.

 

Math 108 - A.

 

I hope I can help others here in time as I get clued up.