1. ## inequality

the question is simplY:

solve for 0<x<10

ln(2x+1)<3 cos x

{it is only woth 4 marks, part one, which was worth 2 marks,was what 2 transformations take
ln(x) to ln(2x+1) so i cant see there is to much work here but i cant get it.}

I can see i need to find the points of intersection, but i must be missing something here.

The mark scheme is no help,it just states the answer.

2. Have you graphed each of these functions?

What do you get?

3. yes,graphed them so i just need points of intersection, but how to find them is the problem

4. Fair problem to have.

You need to use technology (spreadsheet or graphic calculator) to find the intersection of these functions, if you can't use those, a numerical method like the bisection or newton's method will help.

5. Originally Posted by pickslides
Fair problem to have.

You need to use technology (spreadsheet or graphic calculator) to find the intersection of these functions, if you can't use those, a numerical method like the bisection or newton's method will help.
this is what i guessed, but as i say its only 4 marks,compared to the 2 marks for the 2 transformations in part (1).

it gives no starting values and i need 4 numbers (0 being the easy on to find)

on another paper have same sort of question when need to solve e^(-x)-x+1=0, once again mark scheme gives no suggestion as to how the answer is to be found.

6. I can tell you there are 3 points of intersection between $\displaystyle \displaystyle \ln (2x+1)$ and $\displaystyle \displaystyle 3\cos x$ on this interval.

I suggest when using a choosen numerical method, use the starting points as x=1,5 & 7.