Solving a certain equation involving logarithms

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In summary, the conversation is about the book "Prime Obession" which gives an introduction to the Riemann Hypothesis. The author explains that the function log(x) increases more slowly than x raised to any power. The equation log(x) = x^0.1 has two solutions and can be solved using numerical methods or the Lambert W function.
  • #1
bitrex
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I'm reading book called "Prime Obession" which attempts to give a layperson's introduction to the Riemann Hypothesis. In laying the groundwork in one of the early chapters the author is explaining the fact that the function log(x) increases more slowly in total than x raised to any power. For example, log(x) crosses the function x^0.1 somewhere less than e^e, and then falls below it again somewhere in the quadrillions. I'm wondering how one would go about solving the equation to find the Y axis intercepts? I want to find where log(x) = x^0.1, in trying to simplify the problem I end up with log(x)/x = e^0.1, but I'm not able to go any farther. The equation obviously has two solutions, so I imagine there's a polynomial involved somewhere? Any advice would be appreciated.
 
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  • #2
bitrex said:
For example, log(x) crosses the function x^0.1 somewhere less than e^e, and then falls below it again somewhere in the quadrillions. I'm wondering how one would go about solving the equation to find the Y axis intercepts? I want to find where log(x) = x^0.1, in trying to simplify the problem I end up with log(x)/x = e^0.1, but I'm not able to go any farther. The equation obviously has two solutions, so I imagine there's a polynomial involved somewhere? Any advice would be appreciated.

log(x)/x = e^0.1 is not the same as log(x) = x^0.1.

Generally, equations like this have no easy solutions (though they can be solved in terms of a special function called Lambert's W). The best way is probably by numerical methods:

Code:
gp> solve(x=1,9,log(x)-x^.1)
time = 0 ms.
%1 = 3.0597266796208088546065494702258610157
gp> solve(x=1e15,1e16,log(x)-x^.1)
time = 0 ms.
%2 = 3430631121407801.2027753365093892641824
 
  • #3
Yes, you're right. I made a mistake in playing around with the equation: log(x^0.1) != log(x)^0.1 . I considered that numerical methods would probably be a way to solve it; I just wasn't sure if there were some trick that I was missing. Thanks so much for your speedy reply!
 
  • #4
This is equivalent to the idea that any function [itex]f(x)=a^x[/itex] for all values a>1 will rise faster than any polynomial of nth degree such as [itex]g(x)=x^n[/itex] n very large.
 
  • #5
log(x) = x^(0.1)
exact solution is not possible in terms of your standard schoolbook functions
modern CASs like Maple will solve this using the Lambert W function...

x = 10000000000 (W(-0.1))10
 

FAQ: Solving a certain equation involving logarithms

1. How do I solve an equation involving logarithms?

To solve an equation involving logarithms, you need to use logarithmic properties such as the product rule, quotient rule, and power rule. You also need to isolate the logarithm on one side of the equation and use exponentiation to eliminate the logarithm.

2. What is the purpose of using logarithms in equations?

Logarithms are used to solve for unknown variables in equations where the unknown appears in the exponent. They also allow us to condense and simplify complex expressions.

3. Can I solve an equation with multiple logarithms?

Yes, equations with multiple logarithms can be solved by using logarithmic properties to combine the logarithms into a single expression. Then, the same steps as solving a single logarithm equation can be followed.

4. How do I know if my solution to an equation involving logarithms is correct?

You can check your solution by plugging it back into the original equation and simplifying. If the solution satisfies the original equation, then it is correct.

5. Can an equation involving logarithms have extraneous solutions?

Yes, equations involving logarithms can have extraneous solutions, which are solutions that do not satisfy the original equation. It is important to always check for extraneous solutions when solving logarithmic equations.

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