Solve for x (multiplication of trigonometric function and logarithmic function)

brianz24

New member
Joined
Jun 4, 2018
Messages
8
Solve for x in terms of y (trigonometric function x logarithmic function)

y = 1.6log10(x)*cos((4*pi*x)/(4.4+pi))*cos((10*pi*x)/11)

I have this equation from a model that I am trying to implement to my project.
However, the equation is too complex I need help from experts :)
 
Last edited:
y = 1.6log10(x)*cos((4*pi*x)/(4.4+pi))*cos((10*pi*x)/11)

I have this equation from a model that I am trying to implement to my project.
However, the equation is too complex I need help from experts :)
What help do you need - that a calculator cannot give you?!
 
y = 1.6log10(x)*cos((4*pi*x)/(4.4+pi))*cos((10*pi*x)/11)

I have this equation from a model that I am trying to implement to my project.
However, the equation is too complex I need help from experts :)
By 1.6log10(x) do you mean 1.6log10(x) or 1.6log10[10(x)]. In either case, I do not think that you can solve this equation for x in terms of y.
 
By 1.6log10(x) do you mean 1.6log10(x) or 1.6log10[10(x)]. In either case, I do not think that you can solve this equation for x in terms of y.

log10 is log10

The thing is I need to solve for x so I could write the program that calculates the parameter straight-forwardly.

If it is unsolveable, then I might try another method to solve this one.

Please let me know if anyone thinks otherwise.
 
log10 is log10

The thing is I need to solve for x so I could write the program that calculates the parameter straight-forwardly.

If it is unsolveable, then I might try another method to solve this one.

Please let me know if anyone thinks otherwise.
As far as I can tell the original equation cannot be inverted in closed form. The roots can be "approximated" by using different numerical methods (e.g. Newton's method)
 
As far as I can tell the original equation cannot be inverted in closed form. The roots can be "approximated" by using different numerical methods (e.g. Newton's method)

I found Maclaurin series to be an option. Currently working on it.

If there is better solution than this, please do tell. :)
 
Top