Find constants a and b in the function ax^b/ln(x) such that f(1/5)=1 and there is a local min at x=1/5
y = ax^b/lnx
y' = (ax^(b-1) (b*logx - 1))/log^2(x)
since x^b is never zero, we want
b logx = 1
b = 1/ln(1/5) = -0.62133
y = ax^b/lnx
y(1/5) = 1, so
a (1/5)^(-0.62133)/-ln5 = 1
a = -ln5/((1/5)^(-0.62133)) -0.59208
y = -0.59208 x^-0.62133 / lnx
check at
http://www.wolframalpha.com/input/?i=-0.59208+x^-0.62133+%2F+lnx+for+0.1+%3C%3D+x+%3C%3D+0.3
Well, I can try to help you with that! But be warned, my expertise lies more in humor than in math. Let's give it a shot anyway!
Let's start by setting x = 1/5 and f(x) = 1. We have the equation f(1/5) = a(1/5)^b/ln(1/5) = 1.
Now, since the function has a local minimum at x = 1/5, the derivative of the function at x = 1/5 must be equal to zero.
Taking the derivative of f(x), we get:
f'(x) = a * b * x^(b-1)/ln(x) - a * x^b/(x * ln^2(x))
Now, plugging in x = 1/5, we get:
f'(1/5) = a * b * (1/5)^(b-1)/ln(1/5) - a * (1/5)^b/((1/5) * ln^2(1/5))
Hmm... This doesn't seem to be going anywhere mathematically fruitful, does it?
It seems like my mathematical skills aren't quite up to the task. My apologies for not being able to solve this problem for you. You might be better off consulting a proper mathematician who can help you find the values of a and b accurately.
To find the constants a and b in the function f(x) = ax^b/ln(x) such that f(1/5) = 1 and there is a local minimum at x = 1/5, we can follow these steps:
Step 1: Substitute the given values into the function equation
f(1/5) = a(1/5)^b/ln(1/5) = 1
Step 2: Simplify the equation using logarithmic properties
a(1/5)^b/ln(1/5) = 1
a(1/5)^b = ln(1/5)
Step 3: Take the natural logarithm of both sides to eliminate the exponential term
ln(a(1/5)^b) = ln(ln(1/5))
ln(a) + bln(1/5) = ln(ln(1/5))
Step 4: Simplify the logarithmic expression on the right side
ln(a) + bln(1/5) = ln(ln(1/5))
ln(a) - bln(5) = ln(ln(1/5))
Step 5: Since we know that there is a local minimum at x = 1/5, we can differentiate the function and find the derivative equation
f'(x) = abx^(b-1)/ln(x) - ax^b/(xln^2(x))
Step 6: Substitute x = 1/5 into the derivative equation
f'(1/5) = ab(1/5)^(b-1)/ln(1/5) - a(1/5)^b/(1/5*ln^2(1/5))
Step 7: Set the derivative equal to zero to find the critical point
ab(1/5)^(b-1)/ln(1/5) - a(1/5)^b/(1/5*ln^2(1/5)) = 0
Step 8: Simplify the equation
ab(1/5)^(b-1)/ln(1/5) = a(1/5)^b/(1/5*ln^2(1/5))
Step 9: Eliminate the common terms and simplify further
b = 1/(1/5*ln(1/5))
b = -5/ln(1/5)
Step 10: Substitute the value of b into the equation from Step 4
ln(a) - (-5/ln(1/5))*ln(5) = ln(ln(1/5))
Step 11: Simplify the logarithmic expressions
ln(a) + 5ln(5)/ln(1/5) = ln(ln(1/5))
Step 12: Solve for ln(a)
ln(a) = ln(ln(1/5)) - 5ln(5)/ln(1/5)
Step 13: Take the exponent of both sides to solve for a
a = e^(ln(ln(1/5)) - 5ln(5)/ln(1/5))
Finally, we have found the values of constants a and b. Substitute the value of b from Step 9 into the equation of a from Step 13 to get the final values of a and b.
To find the constants a and b in the function f(x) = ax^b/ln(x) given that f(1/5) = 1 and there is a local minimum at x = 1/5, we can follow these steps:
Step 1: Use the given information to form two equations:
- Equation 1: f(1/5) = 1
- Equation 2: f'(1/5) = 0 (to find the local minimum)
Step 2: Substitute the given x-value (1/5) into the function f(x) to form Equation 1:
f(1/5) = a(1/5)^b / ln(1/5) = 1
Step 3: Simplify Equation 1:
a(1/5)^b / ln(1/5) = 1
Step 4: Rearrange Equation 1 to find a(1/5)^b:
a(1/5)^b = ln(1/5)
Step 5: Take the natural logarithm of both sides of the equation to eliminate the logarithm:
ln(a(1/5)^b) = ln(ln(1/5))
Step 6: Apply the logarithmic property to simplify the equation:
ln(a) + b * ln(1/5) = ln(ln(1/5))
Step 7: Let y = ln(a) and z = ln(1/5), which gives:
y + bz = ln(z)
Step 8: Solve for y in terms of z:
y = ln(z) - bz
Step 9: Rewrite y = ln(a), and solve for a in terms of z and b:
a = e^y = e^(ln(z) - bz) = ez * e^(-b)
Step 10: Substitute back into Equation 1:
a(1/5)^b / ln(1/5) = 1
ez * e^(-b) * (1/5)^b / ln(1/5) = 1
Step 11: Rewrite the equation as:
ez * (1/5)^b / ln(1/5) = e^(0) (since any number raised to the power of 0 is 1)
Step 12: Simplify the equation:
ez * (1/5)^b = ln(1/5)
Step 13: We have the equation ez * (1/5)^b = ln(1/5). We can solve this equation numerically or using an approximation method, such as Newton-Raphson, to find the values of z and b.
Note: The specific values of a and b may vary depending on the solution to the equation.