how to solve 12*ln(x) - x^(3/2) = 0

Question

I am attempting to calculate the x-axis intercepts of the curve 12*ln(x) - x^(3/2)

I can obtain the answer but I need to know how to achieve this.

Thanks,

Peter

Answer 1

tri harder

12*ln(x)=x^1.5

12*e^ln(x)=1.5*e^x

12x=1.5*e^x

8x=e^x

ln(8x)=ln(e^x)=x

x=ln(8)+ln(x)

or x-2.0794415416798357  =ln(x)

log(1)=0 for all bases >0

me assume ln(x)>0, impli x>2.0794..., so tri ges=3

ln(3)=1.098612, & 3-2.079...=0.92055845833, pretty kloes

yu kan iterater tu get kloeser

Comments

Second line of your calculation isn't true. 12 needs to be part of the exponent.

Answer 2

This question has been asked by others, Peter, so you're not the only one struggling! It was a good idea to roughly plot the function, but in this tentative solution I've used a simpler plot you might consider.

The solution is going to require trial and error, I believe. I'm hoping to make the solution a bit simpler by first using a go-between variable y=ln(x). You'll appreciate why soon.

We can write 12y=x^(3/2) and take natural logs of both sides:

ln(12y)=(3/2)ln(x)=(3/2)y

So now we have an equation containing one variable y. How to solve ln(12y)=(3/2)y? We use an iterative process by guessing at a value close to the solution. One way to do this is to plot two functions on one graph, one for the left side of the equation and one for the right. When we do this in rough form we get a log curve crossing the y (horizontal) axis at y=1/12 and a straight line passing through (0,0). From the intersection of these two graphed functions we can see that y must be between 2 and 3. That's our starting point. Using a calculator we can get to the next decimal place and discover that y is between 2.1 and 2.2. Then we continue to the next decimal place and discover that y is between 2.17 and 2.18, and so on, until we get y=2.1745 approx., and since y=ln(x), x=e^y, so x=8.7974 approx.

Let's test the answer by substitution. When we compute 12ln(8.7974)-8.7974^(3/2), we get 0.0000244 approx. Pretty close to zero, huh?

Comments

Hi Rod, Thank you very much for your answer and your patients in explaining it.
I hope the inclusion below of the original question which I am working from is permissible but a comment from you about the level of the question would be helpful i.e. am I reading too much into it and making it too complex?

Core Maths for Advanced Level (L. Bostock S. Chandler)
Examimation Questions D
No. 38
(I am unable to include the graph as it is a *.jpg)

The function f is defined for positive real values of x by f(x) = 12*1n(x) – x^(3/2).
The figure shows a sketch of the curve with equation y = f (x). The curve crosses the x-axis at the points A and B. The gradient of the curve is zero at the point C.

a) By calculation, show that the value of x at the point A lies between 1.1 and 1.2.
    The value of x at the point B lies in the interval (n, n + 1), where n is an integer.

b) Determine the value of n.

c) Show that x = 4 at the point C and hence find the greatest positive value of f(x), giving your answer     to 2 decimal places.
d) Write down the set of values of a: for which f (x) is an increasing function of*.    (Edexcel)

I will understand if you would rather not comment and I will be content with you second answer.
Thanks once again.
Regards,
Peter

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When I looked at this again I discovered that there were two solutions; one was between x=1 and 2, and the other between 8 and 9 (the one I found). It has helped having the original question, which has another part, I see. The other solution is point A at x=1.10107 approx. As for the gradient, we differentiate to get 12/x-3sqrt(x)/2, which is zero when 24=3x^3/2, i.e., x^3/2=8 and x^3=64, so x=4, making f(x)=8.64 to 2 dec places. This is a maximum. I'm no expert in judging levels, but I'd say because it includes logs and calculus it would be fairly advanced. Harking back to my youth, I guess I would be doing this sort of question at age 16 or 17, but I don't know what today's standards are. Good to hear from you, Peter. Please keep in touch.