Math Is Fun Forum

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

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

Hi Dave,

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

From

Bob

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

Hi bobbym

I said I'd ask my son whilst in Germany, and he immediately suggested Cauchy's Integral Formula (see Wiki).  From the way you have replied, it sounds like you've thought this too.

The real coefficients, plus the even function mean there must be 3 roots in each quadrant of the complex plane.  He substituted y = x + 1 into one of the order 6 polys and got the coefficients of the other order 6 poly, but in the reverse order.  From this he concluded that if z is one root, then the other two in that quadrant must be 1/(1-z) and -1 + 1/z.  I've still got to get my head around how he reached this conclusion but I'm sure he's right.

He then thinks you can express the integral in terms of z, and conjectures that the zs will go out without having to be evaluated and leaving an imaginary value which, together with Cauchy, will lead to the required result.

But, he also thinks this problem may require one more twist to achieve all I have just stated.  He will probably keep thinking now he's got a taste for the problem and may add his own posts subsequently.

All for now, as I've got some catching up to do on-line.

Bob

See also http://www.bundy.demon.co.uk/Lessons/LessonIndex.htm  choose 'quadratic expressions'

I did this page a while ago.  I think the book references are to Heinemann Intermediate GCSE Mathematics but I've lent the book so cannot check now.

Bob

Hi atut,

I start by looking for factors of 11 (only 11 x 1) and 28 (= 28 x 1 = 14 x 2 = 7 x 4 )

Any of those might be the factorisation so I do a quick check like this:

[ I'm picking one x term to go with one non-x term]

11 x 28          + / -          1 x 1             no good
11 x 1           + / -          1 x 28         also no good

11 x 14          + / -          1 x 2          .........
11 x 2            + / -          1 x 14      ...............

11 x 7             + / -         1 x 4       ...............

but
11 x 4             +              1 x 7       = 51

so (11x +7)(1x + 4)

Here's another example.

2x^2  - 7x    -30

I've got to try 2 x 1 as the x terms and lots of ways to make 30 (30x1=15x2=10x3=5x6)

In other words I'm trying things like (2x -15)(x +2) and (2x + 10)(x - 3)

2    30        2    15        2    10          2    5
1     1         1     2         1     3          1    6

Out of all these I notice that 2 x 6 - 1 x 5 = 7 so that's the one I want.

(2x      5)(x      6)

Now sort out the signs.  One must be a minus and I want -7x so

(2x  +  5)(x  -  6) = 2x^2  - 12x + 5x -30 = 2x^2 -7x -30

The diagram might help.

A 10% increase is the same as multiplying by 1.1 so just do 28 x 1.1 x 1.1 x 1.1=
The formula would be multiplier = 1 + percentage/100     eg. Increase by  25% becomes 1 + 25/100 = 1.25

and decrease by 20% becomes 1 - 20/100 = 0.8

If you've got two sorts of plants and 35% are one type then the rest must be 100 - 35 = 65%.  No trick.

If you do a 'quick post' then all you can do is type some text.

If you click on 'Post reply' you get more options.  Under the text space you must select how many images you want to upload and then you browse your computer to find them on your hard disc.

I often edit my posts after I've sent them because I've spotted a spelling mistake or want to add to an explanation.  There's a button for this too.  But I've never managed to edit a picture once I've sent it.

Hi again,

In the middle of the night I had a further thought and I now think the last lines above need this addition.

(n-a)(n-b)(n-c)(n-d)Q(n) = 0 + 13

It is possible to choose 'n' and 'a' so that (n-a) = 13 (say) and 'b' so that (n-b) = 1 and 'c' so that (n-c) = -1 and Q so that Q(n) = -1, but, as a, b, c, d are said to be distinct, there is no other factor of 13 in the set of integers for (n-d) and therefore ....

This is a contradiction as 13 cannot have these many (distinct) integer factors, so the assumption was false.

Hi DaveRobinsonUk.

I haven't learnt Latex yet so I'm having to explain this just with text, sorry.

If k is a constant (ie. not a variable like x) then it's ok to multply top and bottom by it, and to extract the top one outside the limit, as it isn't dependant on h.

Then I think there's an error, either in your post or in the text because the e in the limit should be to the power 'kh' not just 'h'.

Treat kh as if it was a single variable tending to zero.

Somewhere earler in the text you should have met that this limit  evaluates to 1, which makes the differential k times e to the kx, as required.

Does that help?  If not, I'll have to start learning how to do fancy scripts.

Bob

Hi

Thanks for the factorisation.  The symmetry makes me think this is the way to go but, so far, I've not got anywhere.  On Wednesday 18th I'm off to Germany to visit my son, David.  He's much better at maths than me so I'll ask him to have a try.  Let you know if he comes up with anything.

Question to Heirot.

Where did this problem come from ... maybe there's a clue in that?

Bob

Hi bobbym

Thanks and nice to see you too.  I'm going to track down that nasty integral post and ask a question there.

bob