# : need to solve this question

1. Dec 9, 2005

### sssa

URGENT: need to solve this question

hi ive applied 4 phy at oxf (undergraduate) can u plz tell me how 2 solve question 9...ill be leaving 2morrow, so plz answer quickly..

thanks

Last edited by a moderator: May 2, 2017
2. Dec 9, 2005

### TD

With given y=ln(x) and y=ax it clear that for a certain x, ln(x) = ax. At least, we're looking for such an a.
There is however a second condition in the point, (look at the hint) you can express that as: (ax)' = (ln(x))' <=> a = 1/x.
Now substitute this a = 1/x in the first condition, this yields: ln(x) = 1. For what x does this hold? So what is a?

3. Dec 9, 2005

### Curious3141

Unfortunately, we cannot give you wholesale solutions, so I'll give you a hint.

You need the line to touch the curve - that means that at that value of x, call it 'X' (big X), the y-coordinates are also equal. Come up with one equation to represent this.

Next, you know that the gradient of the curve is the same as that of the line (which is a) at this point. Differentiate the curve to find the gradient at x = X. Now come up with another equation to represent this.

Solve the two simultaneously for a, and you have your result.

Last edited by a moderator: May 2, 2017