Parse error at '<': usage might be invalid MATLAB syntax

% the fluid flow through any of the ipe or chimney can be calculated using
% the realtion of the Moody chart and the Colebrook equation
D = 0:0.1:5;
epsilon = 100:100.5:500;
Re = 2300:2300.5:2900;
f = 0:0.1:10;
i = 0:0.1:1000;
for(i<1100)
    1/((f)^2) = -2.0 log[((epsilon/D)/3.7) + ((2.51/Re*((f)^2)))];

plot(f,D);
title('The Moody chart');

[Ive J]

what about learning MATLAB before using :) ?

mathworks.com/learn/tutorials/matlab-onramp.html

[DGM]

For starters:

% these all have different lengths
% some (e.g. Re) appear to be mistakes
D = 0:0.1:5; % 51 elements
epsilon = 100:100.5:500; % 4 elements
Re = 2300:2300.5:2900; % 1 element
f = 0:0.1:10; % 101 elements
i = 0:0.1:1000; % 10001 elements

% this isn't how a for-loop works
for (i<1100) % this expression is assigned to nothing
    % since nothing in the loop depends on the variable that doesn't exist
    % what is the purpose of the loop?
    
    % this is an invalid assignment with several internal problems
    % the RHS is not a target for assignment.
    % log() is the natural log. colebrook eqn needs log10()
    % use parentheses, not square brackets for function scoping
    % use elementwise operators ./ .^ when dealing with non-scalars
    % since none of the vector lengths match, none of this will work anyway
    1/((f)^2) = -2.0 log[((epsilon/D)/3.7) + ((2.51/Re*((f)^2)))];
% the loop structure is never closed

Given the state of this code, I'm not sure what it's even supposed to be doing.

EDIT: I'm guessing you're trying to solve the equation. You could do that a few ways, but you could also use existing works:

https://www.mathworks.com/matlabcentral/fileexchange/72247-colebrook-white-equation

If nothing else, you can look at how they solve the equation.