Logarithms... :(

[deltro300111]

I've been hunting around for some information to answer my question, but at no avail... as far as understandability goes... I am student in his freshmen year, just finishing Geometry, our teacher has said nothing about the log( function on the calculator, and after expirimenting, I have no idea what it does... I realize there is the e^(pi*i)+1=0 topic already near the top of the page, but I've read through, and found no help.

What are logarithms, what are they used for, and as a geometry student, what would I find them useful for? Keep it a bit lower level, for this is the reason I didn't simply look my answer up on wikipedia, or google... a bunch of scary theorems/postulates I know nothing of. I appreciate any help.

[sleeplessdragn]

RE: Logarithms...

The simplest explanation I have of log is this: it is basically reverse exponents.

Say you are asked to solve for X in this equation: 10^x=100. Using logic, you can deduce that x is equal to 2. But if you were to apply logs in this manner you could rewrite the equation like this: log(base number 10) 100 = x. the log functions calls for a base number, and on most calculators that base number is 10. In fact this base number is so common, that usually it is left out.

Basically log is displaying what the base number to what power will equal what is being "logged". Thus if you type in log(100) you will recieve 2 for an answer, because 10 (the common base number in this situation) to the 2 power is equal to 100.

Basically this is all you need to know for right now, IMO.

[deltro300111]

RE: Logarithms...

Any way to change the log base number? It seems like there would be SOME way to do it. (It's on a Ti-83+)

[The_Q]

RE: Logarithms...

No way to do it on a ti-83. That's stuck on log base ten. You'd have to do the base change formula to get it from a different base to base ten. That formula is.

log(base 10)x/log(base 10) a

Where a is the original base. (Original base being something like log(base 2).

Q

[Brainmaster07]

Unless your math curriculum is a lot different then mine, you shouldn't even need a calculator at all. Just understand the following examples and your good;

(Note that I'm putting the bases in paranthesis because I don't feel like using subscripts)

a) Write the following in logarithmic form

1. b^n = x

2. 5^-2 = 1/25


b) Write the following in exponential form

1. log(2)32 = 5

2. log(6)1/36 = -2


c) Evaluate the following.

1. log(2)16

2. log(8')1 (ignore the ', stupid smilies)


d) What number is n?

1. log(n)1/5 = -1

2. log(2)1/2 = n



a) Compare the exp. form to the log from and just put in the values where appropriate. log(b)x = n is the same as b^n = x.

1. log(b)x = n
2. log(5)1/25 = -2


b) Same thing, just backwards

1. 2^5 = 32
2. 6^-2 = 1/36


c) We know b and x, so solve for n.

1. log(2)16
2 to what power makes 16?
n = 4 (log(2)16 = 4)
4

2. log(8')1
8 to what power makes 1?
n = 0 (log(8')1 = 0)
0

d) Two variables are given, solve for the missing one


1. log(n)1/5 = -1
n^-1 = 1/5
what to -1 equals 1/5?
n = 5

2. log(2)1/2 = n
2^n = 1/2
2 to what is 1/2?
n = -1



That's really it, until you get into anti-logs, natural logs and anti-natural logs. You can also do geometric series with logs... but thats probably not for at least another year.




And you can use the following three log rules to do things like this;

log(b)xy = log(b)x + log(b)y

log(b)x/y = log(b)x − log(b)y

log(b)x^n = nlog(b)x


So now....


Solve for x.

log 2^(3x + 1) = 5
(3x + 1) log 2 = 5
3x log 2 + log 2 = 5
3x log 2 = 5 − log 2
x = (5 − log 2)/(3 log 2)

These can also be bastards (I doubt you will see these for a while);

Solve for X

10^(3x − 1) = 2^(2x + 1)
log10^(3x − 1) = log2^(2x + 1)
(3x − 1) = (2x + 1) log 2
3x − 1 = 2x log 2 + log 2
3x − 2x log 2 = 1 + log 2
x(3 − 2 log 2) = 1 + log 2
x = (1 + log 2)/(3 − 2 log 2)


And yes, I am bored.

[gamefreak721]

i can c u would b after typing that much ..... unless u copied straight of a textbook lol

[banditcom]

i can c u would b after typing that much ..... unless u copied straight of a textbook lol

And I can see you're a fucking moron.

[ckj846]

Umm, that about sums it up. Just remember the rules for logs
log(x)+log(y)=Log(xy)
log(x)-log(y)=Log(x/y)
and so forth. For geometry, this is basically all you need for simplifiying the equations. As I am a sophmore, I can tell you that this is important for Algebra2trig. Graphing logs are simple curves that start on the y/helping axis and curve along the horizontal asymtote.
O_o