How can I solve inequations with logarithms?

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quacam09
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Hi all,

Do you know how to solve the following inequations?

[tex] \ln \left( {\frac{{x + a}} {{x + b}}} \right) \leq cx + d[/tex]
[tex] \ln \left( {\frac{{x + a}} {{x + b}}} \right) \geq \frac{{x^3 + cx^2 + dx + e}} {{ux^2 + vx}} [/tex]

a, b, c, d, e, u, v are constants.
x is a variable.

Can you suggest me a book about solving inequations containing logarithm?
Thanks!
 
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I don't even think it is possible to explicitly solve for x in these two cases. Did you just make these examples up, or you got them from a textbook?
 
quacam09 said:
Hi all,

Do you know how to solve the following inequations?

[tex] \ln \left( {\frac{{x + a}} {{x + b}}} \right) \leq cx + d[/tex]
[tex] \ln \left( {\frac{{x + a}} {{x + b}}} \right) \geq \frac{{x^3 + cx^2 + dx + e}} {{ux^2 + vx}} [/tex]

a, b, c, d, e, u, v are constants.
x is a variable.

Can you suggest me a book about solving inequations containing logarithm?
Thanks!
The standard way of solving complicated inequalities is to solve the corresponding equation first, giving values at which ">" changes to "<", then decide which of the area between those are ">" and which "<".

However, as sutupid math said, there are no "algebraic" methods of solving equations in which the variables appear both in and out of the logarithm (or, more generally, any transcendental functions).
 
HallsofIvy said:
The standard way of solving complicated inequalities is to solve the corresponding equation first, giving values at which ">" changes to "<", then decide which of the area between those are ">" and which "<".

However, as sutupid math said, there are no "algebraic" methods of solving equations in which the variables appear both in and out of the logarithm (or, more generally, any transcendental functions).

Thank you for your response. Are there any method to solve approximately it? Can you give me a suggestion?
 
sutupidmath said:
I don't even think it is possible to explicitly solve for x in these two cases. Did you just make these examples up, or you got them from a textbook?

Thank you.
 
quacam09 said:
Thank you for your response. Are there any method to solve approximately it? Can you give me a suggestion?
Sure. Of course, numeric methods will require that you give specific numerical values to a, b, c, and d. I would recommend "Newton's method". To solve
[tex]ln(\frac{x+a}{x+b}= ln(x+a)- ln(x+b)= cx+ d[/tex],
let [tex]f(x)= ln(x+a)- ln(x+b)- cx- d[/tex]
You want to find x such that f(x)= 0. Choose some starting [itex]x_0[/itex] and then use the iteration
[tex]x_{n+1}= x_n+ \frac{f(x_n}{f'(x_n)}[/tex].
 
HallsofIvy said:
Sure. Of course, numeric methods will require that you give specific numerical values to a, b, c, and d. I would recommend "Newton's method". To solve
[tex]ln(\frac{x+a}{x+b}= ln(x+a)- ln(x+b)= cx+ d[/tex],
let [tex]f(x)= ln(x+a)- ln(x+b)- cx- d[/tex]
You want to find x such that f(x)= 0. Choose some starting [itex]x_0[/itex] and then use the iteration
[tex]x_{n+1}= x_n+ \frac{f(x_n}{f'(x_n)}[/tex].
Thank you.