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C3 help please?

Hi im doing the review exercises for C3 in my text book, and the question is 'show that the equation x=ln(x+5) has a root between x=1 and x=2'. i just cant seem to do it :/ help would be much appreciated!
Reply 1
Original post by noodlemon
Hi im doing the review exercises for C3 in my text book, and the question is 'show that the equation x=ln(x+5) has a root between x=1 and x=2'. i just cant seem to do it :/ help would be much appreciated!


What have you tried?
Original post by noodlemon
Hi im doing the review exercises for C3 in my text book, and the question is 'show that the equation x=ln(x+5) has a root between x=1 and x=2'. i just cant seem to do it :/ help would be much appreciated!


Hint:

Let

f(x)=xln(x+5)f(x)= x-\ln(x+5)

and notice that if, at x=a, f(x) crosses the x axis, it must be above on one side of a and below on the other.
Reply 3
Original post by TenOfThem
What have you tried?

ok ive tried getting rid of the ln getting x+5=e^x? i suppose you need to make x the subject now right?
Reply 4
Original post by noodlemon
ok ive tried getting rid of the ln getting x+5=e^x? i suppose you need to make x the subject now right?


Nope

Did you try putting 1 and 2 in to see what happens
Reply 5
Original post by Indeterminate
Hint:

Let

f(x)=xln(x+5)f(x)= x-\ln(x+5)

and notice that if, at x=a, f(x) crosses the x axis, it must be above on one side of a and below on the other.

so are you saying to use trial and error?
Reply 6
Original post by noodlemon
so are you saying to use trial and error?


You are not trying to find x

RTQ
Reply 7
Original post by TenOfThem
You are not trying to find x

RTQ

Oh i see now, sorry ah good advice.. i subbed in 1 to the equation and got 0.7917... and 2 to get -0.054.. meaning it must have crossed the x axis between those points of 1 and 2. thanks :smile:
Original post by noodlemon
so are you saying to use trial and error?


Well, no.

x=ln(x+5)xln(x+5)=0\displaystyle x=\ln(x+5) \Rightarrow x-\ln(x+5)=0

Thus if you let f(x) equal the LHS, you need to show that there is a root of f(x)=0 between 1 and 2 (i.e it crosses the x axis)

Use the fact that, to cross the x-axis, it must be above the x axis at one point (on one side of the root), and below on the other.

See if you can relate what I'm saying to the values of f(x) when you substitute 1 and 2 in :smile:
Reply 9
Original post by Indeterminate
Well, no.

x=ln(x+5)xln(x+5)=0\displaystyle x=\ln(x+5) \Rightarrow x-\ln(x+5)=0

Thus if you let f(x) equal the LHS, you need to show that there is a root of f(x)=0 between 1 and 2 (i.e it crosses the x axis)

Use the fact that, to cross the x-axis, it must be above the x axis at one point (on one side of the root), and below on the other.

See if you can relate what I'm saying to the values of f(x) when you substitute 1 and 2 in :smile:

yup aha thanks guys :smile: i was overcomplicating the question and trying to find the exact value of x :3

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