How do I solve question 9 on the mathstest?

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SUMMARY

The discussion focuses on solving question 9 from the Oxford undergraduate physics math test, specifically involving the equations y=ln(x) and y=ax. The key steps include finding the value of 'a' such that ln(x) = ax at a specific point, and using the derivative to ensure the line touches the curve. The solution requires setting up two equations: one for the equality of y-coordinates and another for the equality of gradients, which can then be solved simultaneously to find 'a'.

PREREQUISITES
  • Understanding of natural logarithms and their properties
  • Knowledge of differentiation and how to find gradients
  • Ability to solve simultaneous equations
  • Familiarity with the concept of tangents to curves
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  • Study the properties of logarithmic functions and their derivatives
  • Learn how to differentiate functions to find gradients
  • Practice solving simultaneous equations in calculus contexts
  • Explore the concept of tangents and points of tangency in calculus
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This discussion is beneficial for undergraduate physics students preparing for math tests, particularly those focusing on calculus and logarithmic functions. It is also useful for educators and tutors assisting students in understanding these concepts.

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URGENT: need to solve this question

hi I've applied 4 phy at oxf (undergraduate) can u please tell me how 2 solve question 9...ill be leaving 2morrow, so please answer quickly..
http://www.physics.ox.ac.uk/admissions/degrees/questions/mathstest.htm

thanks
 
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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?
 
sssa said:
hi I've applied 4 phy at oxf (undergraduate) can u please tell me how 2 solve question 9...ill be leaving 2morrow, so please answer quickly..
http://www.physics.ox.ac.uk/admissions/degrees/questions/mathstest.htm
thanks

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.
 
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