# How can I solve single nonlinear equation

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Anwar on 23 Jul 2018
Commented: Anwar on 23 Jul 2018

Aarti Dwivedi on 23 Jul 2018
func = @(x) coeff1*x.^exp1 + coeff2*x.exp2 + b;
x0 = 1; % starting point
roots = fzero(func,x0)
In the above code snippet, you can define your function, choose a starting point and fzero will give you the roots. It is generally a good idea to plot your function to get an idea of what it looks like. You can do that in the following manner:
x_interval = -10:10
fplot(func,x_interval)
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Walter Roberson on 23 Jul 2018
There is no real-valued solution for
6.84e24X^1.8164+9.95E13*X+1=0
Consider that if X is positive then all of the terms are positive and so the equation cannot be equal to 0. But if X is negative and real-valued, then you have a negative number raised to a fractional power, which is something that is defined as returning a complex result.
Anwar on 23 Jul 2018
Many thanks, Roberson, for highlighting that, I missed that and was trying to solve in different ways.
Thank you once again.