# MathsWatch

This discussion is closed.
#1
Show graphically or otherwise that the equation x ln x = 1 + x has only one real root and prove that this root lies between 3.5 and 3.8.

TO do the second part I wrote:

x ln x = 1 + x
f(x) = x ln x - x - 1
f(3.5) ~ -0.12
f(3.8) ~ 0.27

f(3.5) < 0 < f(3.8)
Therefore one root lies between 3.5 and 3.8, as the sign of f(x) changes when x goes between these values.

But I don't know how to do the first part. How does the x before ln x and the (-x-1) change the graph (wait a sec, the -1 is obvious but I don't know the x and the -x).

I tried doing:

x ln x = 1 + x
x ln x - x = 1
x (ln x - 1) = 1

so either x or (ln x - 1) equals 1:

x = 1
ln x - 1 = 1
ln x = 2
x = e^2?

but obviously this is wrong...
0
14 years ago
#2
You can't factorise something and set it equal to 1 and say the factors are equal to 1. It only works if something equals 0, because for something to equal zero one of its factors must equal zero. For something to equal 1, its factors may be 2 and 0.5, niether of them are equal to 1.
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#3
(Original post by fishpaste)
You can't factorise something and set it equal to 1 and say the factors are equal to 1. It only works if something equals 0, because for something to equal zero one of its factors must equal zero. For something to equal 1, its factors may be 2 and 0.5, niether of them are equal to 1.
Oh yeah of course, that makes sense. Thanks.
So would you recommend just drawing the diagram?
0
14 years ago
#4
(Original post by mik1a)
Show graphically or otherwise that the equation x ln x = 1 + x has only one real root and prove that this root lies between 3.5 and 3.8.

TO do the second part I wrote:

x ln x = 1 + x
f(x) = x ln x - x - 1
f(3.5) ~ -0.12
f(3.8) ~ 0.27

f(3.5) < 0 < f(3.8)
Therefore one root lies between 3.5 and 3.8, as the sign of f(x) changes when x goes between these values.

But I don't know how to do the first part. How does the x before ln x and the (-x-1) change the graph (wait a sec, the -1 is obvious but I don't know the x and the -x).

I tried doing:

x ln x = 1 + x
x ln x - x = 1
x (ln x - 1) = 1

so either x or (ln x - 1) equals 1:

x = 1
ln x - 1 = 1
ln x = 2
x = e^2?

but obviously this is wrong...

what level math is this???
0
#5
(Original post by KOH)
what level math is this???
P2
0
14 years ago
#6
All you do is draw the graph of y=xlnx ; then y= 1+x , then show there is only 1 intersection
0
14 years ago
#7
(Original post by mik1a)
P2

0
14 years ago
#8
(Original post by mik1a)
Oh yeah of course, that makes sense. Thanks.
So would you recommend just drawing the diagram?
Yeah, I'm just thinking the neatest way to show it, you can see that the rearrangement x = xlnx -1, the RHS is strictly increasing for x > 0 and not defined for x <= 0. So if you were to draw the line y = xlnx - 1, and on the same graph the line y = x, I think you could show that they won't intersect more than once so the equation only has one real root.
0
#9
(Original post by manhunter)
All you do is draw the graph of y=xlnx ; then y= 1+x , then show there is only 1 intersection
(Original post by fishpaste)
Yeah, I'm just thinking the neatest way to show it, you can see that the rearrangement x = xlnx -1, the RHS is strictly increasing for x > 0 and not defined for x <= 0. So if you were to draw the line y = xlnx - 1, and on the same graph the line y = x, I think you could show that they won't intersect more than once so the equation only has one real root.
Of course... because if you solve simultaneously, you get the roots!
Thankyou!
0
14 years ago
#10
man, i cant believe this ****!
0
14 years ago
#11
(Original post by KOH)
man, i cant believe this ****!
Why so worried? Doing P2?
0
14 years ago
#12
(Original post by fishpaste)
Why so worried? Doing P2?

ya, and i thought i had it mastered!!! and i cant even SEE the solotuin to this rather lame question!
0
#13
Ok, new problem..

Show that the equation sin x - ln x = 0 has a root lying between x = 2 and x = 3. Given that this root lies between a/10 and (a+1)/10, where a is an integer, find the value of a.

I know this is basically asking me to find the root between two numbers to a decimal place, so a is obviously between 20 and 30. I could probably work this out by trial and error, but that's not the right way I don't think.

I drew y = ln x and y = sin x and they do cross ar roughtly the right point, but how do I find this point accurately? This question is before the section explaining the iterative method, so I guess I would need to find it out by some other way? (besides, how on earth do you rearrange sin x - ln x = 0 into the correct form... x = arcsin(ln x) ?)

Thanks for any help!
0
14 years ago
#14
(Original post by KOH)
ya, and i thought i had it mastered!!! and i cant even SEE the solotuin to this rather lame question!
Right the first bit.

xln x = 1 + x

you've got to show that that equation has only one root

A good trick is as manhunter said, to draw y = 1 + x and y = xln x

You should be able to sketch those without too much difficultly.

if y = 1 + x and y = xln x then 1 + x = xln x, ie. the equation we're trying to solve. So the simultaneous equations y = 1 + x and y = xln x should have the same root(s). You can show that these only have one root graphically from your sketch. because y = 1 + x is a straight line, and y = xln x is an increasing function.

So...which bits are you lost on?
0
14 years ago
#15
(Original post by fishpaste)
Right the first bit.

xln x = 1 + x

you've got to show that that equation has only one root

A good trick is as manhunter said, to draw y = 1 + x and y = xln x

You should be able to sketch those without too much difficultly.
if you cant, dont worry, p2 lets you use graphical calculators
0
14 years ago
#16
(Original post by fishpaste)
Right the first bit.

xln x = 1 + x

you've got to show that that equation has only one root

A good trick is as manhunter said, to draw y = 1 + x and y = xln x

You should be able to sketch those without too much difficultly.

if y = 1 + x and y = xln x then 1 + x = xln x, ie. the equation we're trying to solve. So the simultaneous equations y = 1 + x and y = xln x should have the same root(s). You can show that these only have one root graphically from your sketch. because y = 1 + x is a straight line, and y = xln x is an increasing function.

So...which bits are you lost on?
thnks for da help, BUT i had already gotten it, what i meant is that i am worried how i didn come up with the way by my self, thanks for da help anyway, sorry 4 misunderstanding!
0
14 years ago
#17
(Original post by KOH)
thnks for da help, BUT i had already gotten it, what i meant is that i am worried how i didn come up with the way by my self, thanks for da help anyway, sorry 4 misunderstanding!
Ohhh okay. =_^
0
#18
(Original post by mik1a)
Ok, new problem..

Show that the equation sin x - ln x = 0 has a root lying between x = 2 and x = 3. Given that this root lies between a/10 and (a+1)/10, where a is an integer, find the value of a.

I know this is basically asking me to find the root between two numbers to a decimal place, so a is obviously between 20 and 30. I could probably work this out by trial and error, but that's not the right way I don't think.

I drew y = ln x and y = sin x and they do cross ar roughtly the right point, but how do I find this point accurately? This question is before the section explaining the iterative method, so I guess I would need to find it out by some other way? (besides, how on earth do you rearrange sin x - ln x = 0 into the correct form... x = arcsin(ln x) ?)

Thanks for any help!
Back up to the top you go!
0
14 years ago
#19
(Original post by mik1a)
I know this is basically asking me to find the root between two numbers to a decimal place, so a is obviously between 20 and 30. I could probably work this out by trial and error, but that's not the right way I don't think.
Why not? It's the way I'd do it...
0
#20
(Original post by Squishy)
Why not? It's the way I'd do it...
Hmm ok
Thanks.
0
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