# Thread: napierian logarithm problem

1. ## napierian logarithm problem

the function f is defined for positive real values of x by f(x)=12lnx - x^(3/2)
the curve crosses the x-axis at points A and B
a. by calculation, show that the value of x at point A lies between 1.1 and 1.2
b. the value of x at the point B lies in the interval (n, n+1) where n is an integer. find n

for a. i took 12lnx = x^(3/2) and turned it all around, getting e^x^(3/2) = x^12 etc. but i still have no idea of how to get the interval for x
for b. i have no idea

2. Originally Posted by furor celtica
the function f is defined for positive real values of x by f(x)=12lnx - x^(3/2)
This is $f(x)= 12 ln(x)- x^{3/2}$, not $f(x)= 12 ln(x- x^{3/2})$?

the curve crosses the x-axis at points A and B
a. by calculation, show that the value of x at point A lies between 1.1 and 1.2
b. the value of x at the point B lies in the interval (n, n+1) where n is an integer. find n

for a. i took 12lnx = x^(3/2) and turned it all around, getting e^x^(3/2) = x^12 etc. but i still have no idea of how to get the interval for x
for b. i have no idea
What is f(1.1)? What is f(1.2)? Are they of different sign? What does that tell you?

For (b), Calculate f(n) for n= 2, 3, 4, 5,etc. until you find a root!

3. can ou please show me how to work through a.? is there another way of getting b.?
its (12lnx) - (x^(3/2))
what does 'are they of different sign' supposed to mean? i wouldnt be posting this if it was so simple for me, ive thought over it ok?