Naturual Logarithms

DaveRobinsonUK · 2010-08-22 22:36:11

Hi All

I have been stuck on this for a few days now, and seem to be getting more and more perplexed.

It was all going well, and then I hit a few questions that I don't know how to do and can't seem to
find an explanation. Or maybe I have seen it and then forgotten that it applies in this case.

The questions are of the form

Solve:

I am also not seeing how the identities

and work.

Is

the same as based on

Quite confused!

Thank you

bobbym · 2010-08-22 22:43:33

Hi Dave;

I do these like this:

Take e^x of both sides:

Should be able to handle it now.

Bob · 2010-08-22 22:45:39

Hi DaveRobinsonUK,

y = lnx is the inverse function of y = e^x. Your identities are just alternative ways of saying that.

For your question the 'third law of logarithms' allows you to re-write 2ln(x-1) as ln((x-1)^2).

So do that and then 'unlog' the expressions ... strictly raise e to the power of each expression to land up with

(x+1) = (x-1)^2.

Now you should be able to re-arrange this as a quadratic and solve.

Remember that ln Y is only defined in real numbers if Y > 0, so one of your values of x may solve the quadratic but not the original log equation.

Bob

DaveRobinsonUK · 2010-08-22 23:33:58

Hi Bobby.. Hows things pal?
and hello Bob Bundy, thank you for the help.

I got the first one. Though I am still stuck on the next, and the one after that seems to be similar if opposite.

I have done this, though the answer should be

using first law of logs

though I know this can't be right because raising by e^x, just gives the factors given to start with.

bobbym · 2010-08-22 23:36:03

Hi Dave;

How are you doing? You are right that is the correct answer. Move the log(x-1) to the right and take e^x of both sides.

DaveRobinsonUK · 2010-08-22 23:49:46

All is ok here,apart from our neighbors are rowing periodically at the moment. 11PM they were at it last night, sounded like they were throwing pans and plates at each other. Maybe there is someway to calculate when this is going to happen. Then we can arrange to be out.

O.k

so I move ln(x-1) to the right giving

take both sides to e^x

not sure how to get

from that

Feel like I have gone backwards at the moment, do you ever get that?

bobbym · 2010-08-22 23:52:36

Hi Dave;

Please don't get involved with your neighbors. Hope they move!

This is an error:

You lost the minus sign. That move of bringing the minus through the log is illegal. Check what I did in only if you need it.

Feel like I have gone backwards at the moment, do you ever get that?

All the time! Don't think about it. There are always problems that will make you look like a monkey!

Backwards is forward motion in reverse.

Bob · 2010-08-23 00:01:28

Hi Dave,

Looks like you've tried to multiply the minus inside the log function. You cannot do that.

From

you should go to

Bob

Last edited by Bob (2010-08-23 00:02:36)

Bob · 2010-08-23 00:08:31

But the answer to that isn't

????

I've now solved the quadratic (bobbym had the same in post 5 ) and both x values lead to impossible ln(negative).

Please check the original question.

Bob

Last edited by Bob (2010-08-23 00:16:18)

DaveRobinsonUK · 2010-08-23 00:15:18

just to clarify

so when you say raise to e^x

does this mean

bobbym · 2010-08-23 00:15:49

Nope!

That's what e^ to both sides means.

The e^x is the inverse operator for ln and e^(-a) = 1 / e^a, so:

Multiply by x - 1 and you are on your way!

Bob · 2010-08-23 00:32:38

But this problem has no 'ln' solutions.

See picture below taken from Excel.

It shows values of x from -5 to 5.

The error message occurs because ln is not defined in real numbers when x is not positive.

I'm suspecting the 'book' problem didn't look quite like this

CORRECTION. SILLY ME, i PUT THE WRONG NUMBER IN MY CALCULATOR.

ONE SOLUTION OF THE QUADRATIC LEADS TO AN IMPOSSIBLE VALUE. THE OTHER WORKS AS REQUIRED.

Bob

Last edited by Bob (2010-08-23 00:44:07)

bobbym · 2010-08-23 00:40:15

Hi bob;

Here is the root at 1.23606 approximately. √5 - 1 exactly.

http://www.mathsisfun.com/graph/functio … 6666666667

Bob · 2010-08-23 00:46:05

Yes, I award myself twenty imaginary lashes.

Bob

bobbym · 2010-08-23 00:48:24

Hi bob;

Good thing about working with computers, they forgive all our errors!

Now the only remaining question is what happened to Dave! He was here.

DaveRobinsonUK · 2010-08-23 01:01:00

hi guys.

Sorry I had to nip out to feed the toddler some lunch, and also to do a bit of backtracking through the books
back now

bobbym · 2010-08-23 01:03:03

Okay, what was I talking about?

Bob · 2010-08-23 01:04:27

You have a computer that forgives you your errors!

Mine is always telling me off even when it's at fault.

And then when I re-start (this being the only option) it tells me off again for not closing down properly.

I'd like one like yours.

As for Dave, maybe he's still trying to take it all in ... that was a lot of posts in a short time.

Bob

bobbym · 2010-08-23 01:06:43

Oh, okay, thanks! Nah, he had to go feed a toddler. Not really, my computer is ill behaved also. I am hungry to.

Bob · 2010-08-23 01:17:18

By the way, another way to solve hard maths is to have a son who is very good at it. If you haven't got one, it's a rather slow method, but if you have, then all you have to do is fly to Germany and visit him.

I've put the results of his brain power on http://www.mathisfunforum.com/viewtopic.php?id=14345

And, unlike my computer, he doesn't complain.

Bob

bobbym · 2010-08-23 01:23:27

It is a good to have someone to help you.

DaveRobinsonUK · 2010-08-23 01:27:06

I think i see now.

So ln is the reciprocal of e^x.

Here's hoping.

bobbym · 2010-08-23 01:41:09

The inverse operator. Think of it as undoing the other one. Like squaring and the square root. Not rigorous but...

DaveRobinsonUK · 2010-08-23 01:52:43

ok so its a reflection in the line y=x

bobbym · 2010-08-23 02:12:04

Hi Dave;

Don't think of it that way. If you plot e^x and ln(x) on the same scales you will see that they are not mirror images. Like x^2 and sqrt(x) are not. Think of it as undoing the other one.

Math Is Fun Forum

#1 2010-08-22 22:36:11

Naturual Logarithms

#2 2010-08-22 22:43:33

Re: Naturual Logarithms

#3 2010-08-22 22:45:39

Re: Naturual Logarithms

#4 2010-08-22 23:33:58

Re: Naturual Logarithms

#5 2010-08-22 23:36:03

Re: Naturual Logarithms

#6 2010-08-22 23:49:46

Re: Naturual Logarithms

#7 2010-08-22 23:52:36

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#8 2010-08-23 00:01:28

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#9 2010-08-23 00:08:31

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#12 2010-08-23 00:32:38

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