More math!! Please help!

[SkyHog]

Just in time for finals, Nick's completely lost.

I hate logarithms with my entire being. The last test I took was nothing but logarithms, and I got a 22%. Luckily, the teacher drops the lowest score, but I still hate to have a 22% on anything. Turns out the final has a lot of logs. Yay!

Here's a few questions that I have no idea how they work:

First off, we covered a problem today, and part of it worked out this way:

ln(e3r)

I understand that ln(e) = 1, so then that turns into 13r

That turns into just 3r.

But the logic part of my brain says that 1 to any power is just 1. So how the heck does 13r become 3r???

I have a bunch more questions too, but I have to organize my thoughts, and that ain't easy after working graveyards and having my brain blown.

I hate logarithms.

 

[wbarnhill]

Remember that log base a of x equals y, and a^y = x.

and log base e = natural log. (ln)

So therefore

log base e of e^3r = y

e^y = e^3r

So what's y?

 

[Arnold]

This was so wrong I removed it.
Sorry for any misleading I did.

 

[SkyHog]

Remember that log base a of x equals y, and a^y = x.

and log base e = natural log. (ln)

So therefore

log base e of e^3r = y

e^y = e^3r

So what's y?

well, I'd assume that the e's would cancel out, so you'd get y= 1]3r and you're back to my original question....how can 1 to the anything be anything but 1? I know the answer is 3r, but I can't grasp why.

 

[ScottM]

ln (natural log) is the reverse (antilog) of anything that is e^x

What is happening is that you need to perform the operation of e^3r first then ln(x) reverses that operation so you just end up with the exponant.

You are not first taking ln(e) = 1 and then doing 1^3r which of course would be 1.


Another example
Using base 10 algorithms you know that 10^2 = 100

Log (10^2) = Log (100)

What exponent of 10 will give us 100? Answer 2

therefore Log(100)=2

Or your could break up the exponentail like this

Log(10^2) = Log (10^(1+1)) = Log(10 * 10) = Log (10) + Log (10) = 1+1 =2

See it?

A log is asking the question "what exponant must I use for the base to get a number equal to x"

In your equation that translates into what exponant must I use with e to get e^3r? well of course the answer is 3r

Scott

 

[SkyHog]

We are talking some extremely serious rust here, so don't take my word for anything but I think I know - someone please check me on this.

I don't know why e=1 but I'll let that be.

To me the expression
ln(e3r) if I remember this correctly (disclaimer #2) is simplified as
(3r)(e). As I recall, don't quote me (disclaimer #3) the when you simplify or solve a log expression, the exponent becomes whatever you call one of the two thingies you multiply together in a multiplication problem. So assuming e=1 then (3r)(1) = 3r.

Where I think you missed a step is that substituting 1 for e leaves you with
ln(13r) which properly simplifies to 3r * 1.

what I think you did was simplify the expresion 13r first, <resulting in 1> then you are trying to solve the log problem - this leave 0, again if my memory is not too rusty (disclaimer #4) as the log of a constant is always 0, if memory serves . . . (disclaimer #5).

I sure hope I didn't make this worse for you.

Someone please correct me if I'm wrong.

That makes a load of sense. Thank you.

and, btw, ln(e) always =1. Why? Because my teacher said so I forgot about that rule that says that you can move the exponent to the front of the log.

But.....logically, it still doesn't make sense. 1 to any power is always 1, right?

 

[ScottM]

We are talking some extremely serious rust here, so don't take my word for anything but I think I know - someone please check me on this.

I don't know why e=1 but I'll let that be.

To me the expression
ln(e3r) if I remember this correctly (disclaimer #2) is simplified as
(3r)(e). As I recall, don't quote me (disclaimer #3) the when you simplify or solve a log expression, the exponent becomes whatever you call one of the two thingies you multiply together in a multiplication problem. So assuming e=1 then (3r)(1) = 3r.

Where I think you missed a step is that substituting 1 for e leaves you with
ln(13r) which properly simplifies to 3r * 1.

what I think you did was simplify the expresion 13r first, <resulting in 1> then you are trying to solve the log problem - this leave 0, again if my memory is not too rusty (disclaimer #4) as the log of a constant is always 0, if memory serves . . . (disclaimer #5).

I sure hope I didn't make this worse for you.

Someone please correct me if I'm wrong.

Very rusty and very wrong , see my post above for an explaination.

e is NOT equal to one, it is an transcendental number = to 2.71828...

The ln(e) is equal to 1

 

[ScottM]

That makes a load of sense. Thank you.

and, btw, ln(e) always =1. Why? Because my teacher said so I forgot about that rule that says that you can move the exponent to the front of the log.

But.....logically, it still doesn't make sense. 1 to any power is always 1, right?

ln(e)=1 because the log base e of e is 1. natural logs are shortened to ln instead of Loge jsut as log base 10s are just Log, but it is entirely poosible to have Logbase 2 as Log2.

See my post above for the explination

 

[SkyHog]

ln (natural log) is the reverse (antilog) of anything that is e^x

What is happening is that you need to perform the operation of e^3r first then ln(x) reverses that operation so you just end up with the exponant.

You are not first taking ln(e) = 1 and then doing 1^3r which of course would be 1.


Another example
Using base 10 algorithms you know that 10^2 = 100

Log (10^2) = Log (100)

What exponent of 10 will give us 100? 2

therefore Log(100)=2

See it?

Scott

Maybe the whole problem would be easier to see.

Its one of those fancy A=Pert problems, where p=100 when t=0 and A=1800 when t=3 (word problem gives that information).

so then:
1800=100er3
18=er3
ln(18)=ln(e)r3

Now at this point, I get a bit lost. I assume its like this:
ln(18)=r3 <---(the questionable part above)


but then, the final answer works out to:

r = ln(18) / 3


Where does that come from?

 

[Areeda]

Or your could break up the exponentail like this

Log(10^2) = Log (10^(1+1)) = Log(10 + 10) = Log (10) + Log (10) = 1+1 =2

Not quite Log(10^(1+1)) = Log(10 * 10) = Log(10) + Log(10) = 2;

 

[ScottM]

Maybe the whole problem would be easier to see.

Its one of those fancy A=Pert problems, where p=100 when t=0 and A=1800 when t=3 (word problem gives that information).

so then:
1800=100er3
18=er3
ln(18)=ln(e)r3

Now at this point, I get a bit lost. I assume its like this:
ln(18)=r3 <---(the questionable part above)


but then, the final answer works out to:

r = ln(18) / 3


Where does that come from?

So far so good look again at your finaly equation of r = ln(18) / 3

Remeber you said that ln(18) = r3 so substitute

r = r3/3 which of course equals r therefore you just proved you understand the concept.

To get a number you need to go to the log table and find the ln(18) and then put that number in and divide by three. ln(18) from my calculator is 2.8904 divide that by 3 = .9635 is what r is.

 

[SkyHog]

So far so good look again at your finaly equation of r = ln(18) / 3

Remeber you said that ln(1) = r3 so substitute

r = r3/3 which of course equals r therefore you just proved you understand the concept.

hey now...I get it...sort of.

The question is how do I get to the final step where it proves itself like that? There's a logic leap of some sort that goes from:

ln(18)=r3

to

r = r3 /3

 

[ScottM]

Not quite Log(10^(1+1)) = Log(10 * 10) = Log(10) + Log(10) = 1;

ooops!! You are quit right, I changed it up above.

 

[Steve]

Lotta help on the web...

http://www.sosmath.com/algebra/logs/log1/log1.html

http://en.wikipedia.org/wiki/Natural_logarithm

http://www-math.mit.edu/~djk/18_01/chapter04/section04.html

 

[ScottM]

hey now...I get it...sort of.

The question is how do I get to the final step where it proves itself like that? There's a logic leap of some sort that goes from:

ln(18)=r3

to

r = r3 /3

That is just substitution if ln(18) = r3 and you also have the equation r = ln(18)/3 you can replace the term ln(18) with r3, basic algebra. But that is just to prove you did the problem right. At the point where you have numbers all on one side I would simplifiy it to get an answer by doing the log look up (calculator or table) and the division

 

[Areeda]

Maybe this will be more confusing than helpful but I hope not.

How does a Whiz Wheel work?
By adding and subtracting logarithms.

The two disks are labeled logarthmically. When you line up the arrow and read off a numbre you are adding, when you line up the number and read off the arrow subtracting.

This is based on the law of logarithms:

Log(a) + Log(b) = log(ab)
Log(a) - Log(b) = log(a/b)

also (but not on E6B):

a * Log(b) = log(b^a)
Log(b)/a = log(b^-a)

I did find a web reference: http://oakroadsystems.com/math/loglaws.htm

Joe

 

[SkyHog]

That is just substitution if ln(18) = r3 and you also have the equation r = ln(18)/3 you can replace the term ln(18) with r3, basic algebra. But that is just to prove you did the problem right. At the point where you have numbers all on one side I would simplifiy it to get an answer by doing the log look up (calculator or table) and the division

Okie....so then, where did r = ln(18) / 3 come from?

I get the substitution part, but somehow, I'd have to know that final equation.

 

[SkyHog]

Maybe this will be more confusing than helpful but I hope not.

How does a Whiz Wheel work?
By adding and subtracting logarithms.

The two disks are labeled logarthmically. When you line up the arrow and read off a numbre you are adding, when you line up the number and read off the arrow subtracting.

This is based on the law of logarithms:

Log(a) + Log(b) = log(ab)
Log(a) - Log(b) = log(a/b)

also (but not on E6B):

a * Log(b) = log(b^a)
Log(b)/a = log(b^-a)

I did find a web reference: http://oakroadsystems.com/math/loglaws.htm

Joe

Thats really good info to know. Thanks Joe.

 

[ScottM]

Okie....so then, where did r = ln(18) / 3 come from?

I get the substitution part, but somehow, I'd have to know that final equation.

from you, emphasis mine.

I just wanted to show you had done the algebra correctly.

Maybe the whole problem would be easier to see.

Its one of those fancy A=Pert problems, where p=100 when t=0 and A=1800 when t=3 (word problem gives that information).

so then:
1800=100er3
18=er3
ln(18)=ln(e)r3

Now at this point, I get a bit lost. I assume its like this:
ln(18)=r3 <---(the questionable part above)


but then, the final answer works out to:

r = ln(18) / 3


Where does that come from?

 

[SkyHog]

from you, emphasis mine.

I just wanted to show you had done the algebra correctly.

[/I]

Holy crap - it just clicked.

since I got ln(18)=r3 then algebraically....(and in baby step math)

divide both sides by 3 to get ln(18) /3 = r

Reorganized:
r = ln(18) / 3

earlier in the problem, I determined that ln(18) = r3 so:

r = r3 / 3

Wow. This is crazy stuff.

 

[ScottM]

Maybe this will be more confusing than helpful but I hope not.

How does a Whiz Wheel work?
By adding and subtracting logarithms.

The two disks are labeled logarthmically. When you line up the arrow and read off a numbre you are adding, when you line up the number and read off the arrow subtracting.

This is based on the law of logarithms:

Log(a) + Log(b) = log(ab)
Log(a) - Log(b) = log(a/b)

also (but not on E6B):

a * Log(b) = log(b^a)
Log(b)/a = log(b^-a)

I did find a web reference: http://oakroadsystems.com/math/loglaws.htm

Joe

Yep that is how they are used to do more complex math problems. They were very handy precalculator days, not as much anymore.

BTW here is one basic relationship for logs and their expontial forms

logbx = n means bn = x.

b is the base of the log series could be any number

 

[Areeda]

so then:
1800=100er3
18=er3
ln(18)=ln(e)r3

Not quite Nick:
the blue step should read:
ln(18) = ln(e) * 3r

then:

ln(18) = 1 * 3r
ln(18)/3 = r
2.89/3 = r
0.963 = r

As Scott said.

Joe

 

[ScottM]

Holy crap - it just clicked.

Just wait a little longer when you hit calculas it all clicks and all of a sudden math makes perfect sense and becomes a language to describe the world around you.

 

[ScottM]

Not quite Nick:
the blue step should read:
ln(18) = ln(e) * 3r

Can't do that

It is ln(18)=ln(e^r3)

The right side of the equation can be simplified.

The ln(e^r3) is just r3 not ln(e)*r3

 

[Areeda]

Not quite Nick:
the blue step should read:
ln(18) = ln(e) * 3r

Can't do that

It is ln(18)=ln(e^r3)

The right side of the equation can be simplified.

The ln(e^r3) is just r3 not ln(e)*r3

Actually Scott
r3 =ln(e) * r3 = 1 * r3

I think one of Nick's misconceptions from the first post was that ln(ab) = ln(a)b when in reality it is ln(a) * b. I was just trying to emphasize that.

Joe

 

[SkyHog]

I just looked at a few more of the problems on the final review. Thanks for the help guys, but I think I'm gonna have to get a tutor for this. This is just a bit over my head.

Just think - only 3 more years of what can be classified as a very math heavy degree. I won't lose my mind

 

[SkyHog]

actually...I'll try one more - tell me if I'm right:

Suppose that $95,000 is invested at 6% interest, compounded continuously. How long will it take for the amount in the account to reach $12,000?

so....again A=Pert

12000 = 9500e.06t
24/19=e.06t
ln(24/19)=ln(e).06t
ln(24/19)=.06t
ln(24/19) / .06 = t
t = ln(24/19) / .06 years

Is that right? (teacher specifically says not to give a decimal answer)

 

[Areeda]

t = ln(24/19) / .06 years

Is that right? (teacher specifically says not to give a decimal answer)

That's what I get.

Just to verify we want about a 26% increase (12,000/9,500) at 6% (not compounded) that would come out to a little over 4 years and compounded continuously based on our result is 3.89 years. Looks good to me.

Joe

 

[ScottM]

actually...I'll try one more - tell me if I'm right:

Suppose that $95,000 is invested at 6% interest, compounded continuously. How long will it take for the amount in the account to reach $12,000?

so....again A=Pert

12000 = 9500e.06t
24/19=e.06t
ln(24/19)=ln(e).06t
ln(24/19)=.06t
ln(24/19) / .06 = t
t = ln(24/19) / .06 years

Is that right? (teacher specifically says not to give a decimal answer)

Looks right.

Or write it as [100*ln(24/19)]/6 to clean up the decimal in the denominator



For fun: that all equals 3.8936 years

.8936 years * 365day/year = 326.164 days

.164 days * 24 hours/ day = 3.936 hours

.936 hours * 60minutes/hour = 56.16 minutes

.16 minutes *60seconds/minute = 9.6 seconds (go ahead and round up 4/10th of a second

So the non decimal answer is 3 years 326 days 3 hours 56 minutes and 10 seconds

 

[Areeda]

So the non decimal answer is 3 years 326 days 3 hours 56 minutes and 10 seconds

And I thought I was wierd

 

[ScottM]

And I thought I was wierd

Nope it is a large club, we have jackets too!!!
http://www.geek.com/news/geeknews/2003Nov/bpd20031119022611.htm

 

[TMetzinger]

actually...I'll try one more - tell me if I'm right:

Suppose that $95,000 is invested at 6% interest, compounded continuously. How long will it take for the amount in the account to reach $12,000?

so....again A=Pert

12000 = 9500e.06t
24/19=e.06t
ln(24/19)=ln(e).06t
ln(24/19)=.06t
ln(24/19) / .06 = t
t = ln(24/19) / .06 years

Is that right? (teacher specifically says not to give a decimal answer)

95,000 will only change to 12,000 if you LOSE money - everyone else has assumed a typo and the original amount was 9500, but I think you invested in Enron.

 

[ScottM]

95,000 will only change to 12,000 if you LOSE money - everyone else has assumed a typo and the original amount was 9500, but I think you invested in Enron.

He fixed it int he equation, 12000 = 9500e^.06t

But if it was Enron he would have started with $95,000 and ended up with a big goose egg

 

[ScottM]

FYI: Enjoy!

History of Logarithms:
Logarithms were invented independently by John Napier, a Scotsman, and by Joost Burgi, a Swiss. The logarithms which they invented differed from each other and from the common and natural logarithms now in use. Napier's logarithms were published in 1614; Burgi's logarithms were published in 1620. The objective of both men was to simplify mathematical calculations. Napier's approach was algebraic and Burgi's approach was geometric. Neither men had a concept of a logarithmic base. Napier defined logarithms as a ratio of two distances in a geometric form, as opposed to the current definition of logarithms as exponents. The possibility of defining logarithms as exponents was recognized by John Wallis in 1685 and by Johann Bernoulli in 1694.
The invention of the common system of logarithms is due to the combined effort of Napier and Henry Biggs in 1624. Natural logarithms first arose as more or less accidental variations of Napier's original logarithms. Their real significance was not recognized until later. The earliest natural logarithms occur in 1618.

 

[SkyHog]

Thanks guys. I like breaking it down into actual numbers. If I do that on the test my teacher would rip his hair out (I'm sure he doesn't have that answer on his key).

 

[Keith Lane]

Hey Nick, Here's a site I use from time to time.

http://www.mathwords.com/

I like the way they make words in a definition or an example a link. So if you don't understand a definition just keep on drilling down till you're really confused.