Solving equation involving a variable and its logarithm

In summary, the equation ln(x^(3/2))-bx-c=0 does not have an analytical solution and can only be solved numerically. It is the intersection of ln(x) and the line 2(bx+c)/3. The Lambert W function may be considered an analytical solution, but it will only provide real solutions for certain values of n.
  • #1
JulieK
50
0
Can you suggest a general analytical solution to the following equation

[itex]\ln(x^{3/2})-bx-c=0[/itex]

where [itex]x[/itex] is real positive variable and [itex]b[/itex] and [itex]c[/itex] are real positive
constants.
 
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  • #2
There is no analytical solution. It should be solved numerically.
 
  • #3
Sure - it is the intersection of ##y=\ln(x)## with the line ##y=2(bx+c)/3##.
Note: ##\ln(x)## is only defined for ##x>0##.

Solutions will not exist for all a and b.
In general you'll need a numerical solution.
 
  • #4
Shyan said:
There is no analytical solution.
You're probably right, but ... can you prove it?
 
  • #5
Depends on whether or not you classify the Lambert W function as an analytical solution (I doubt the mathematics community does).

$$\ln(x^{\frac{3}{2}}) = b x + c \\
x = \exp( \frac{2}{3}bx + \frac{2}{3}c) \\
x \exp(- \frac{2}{3} bx) = \exp( \frac{2}{3}c) \\
- \frac{2}{3} b x \exp(- \frac{2}{3}bx) = -\frac{2}{3} b \exp( \frac{2}{3}c) \\
- \frac{2}{3} b x = W_n(-\frac{2}{3} b \exp( \frac{2}{3}c)) \\
x = - \frac{3 W_n(-\frac{2}{3} b \exp( \frac{2}{3}c))}{2 b}$$

For a real [itex]x > 0[/itex], [itex]b \neq 0[/itex] and [itex]n \in \mathbb{Z} [/itex]. Only [itex]n=-1, 0[/itex] can provide real solutions, though.
 
Last edited:
  • #6
The W-function is certainly an analytic function so I would call that an "analytic solution".
 
  • #7
Looks like confusion between "analytic function" and "analytic expression".
 

1. What is a logarithm?

A logarithm is a mathematical function that represents the number of times a base number needs to be multiplied by itself to reach a given number. In other words, it is the inverse function of exponentiation.

2. How do I solve an equation involving a variable and its logarithm?

To solve an equation involving a variable and its logarithm, you need to isolate the logarithm term on one side of the equation and then use the logarithm rules to simplify it. Once the logarithm term is simplified, you can solve for the variable using algebraic methods.

3. What are the basic logarithm rules?

The basic logarithm rules are:

  • The logarithm of a product is equal to the sum of the logarithms of the individual factors.
  • The logarithm of a quotient is equal to the difference of the logarithms of the individual terms.
  • The logarithm of a power is equal to the exponent multiplied by the logarithm of the base.
  • The logarithm of the base itself is equal to 1.

4. What is the relationship between logarithms and exponential functions?

Logarithms and exponential functions are inverse functions of each other. This means that the logarithm of a number is the exponent that the base needs to be raised to in order to get that number. For example, log2(4) = 2 because 22 = 4.

5. Why do we use logarithms in equations?

Logarithms are useful in equations because they allow us to solve for the value of a variable that appears as an exponent. This is particularly helpful in situations where the exponent is very large or very small, making it difficult to solve using traditional algebraic methods.

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