How to code a function for solving equation by iteration?

I'd like to write a function which calculates the pressure drop given by:

P = (1/2)*rho*omega^2*(L/D)*lambda

where rho - fluid's density, omega - fluid's velocity, L - pipe's length, D - pipe's diameter, lambda - function of the Reynolds number.

• For Reynolds<2320, lambda = 64/Reynolds
• For 2320<Reynolds<10^5, lambda = 0.3164/(Reynolds^0.25)
• For 10^5<Reynolds<10^6, 1/sqrt(lambda) = 2*log(Reynolds*sqrt(lambda)) - 0.8, to be solved via iteration according to this algorithm: letting tol = 0.01 and lambda0 = 1, the starting value x0 should be x0 = 1/sqrt(lambda0) whereas the next value is x = 2lg(Reynolds/x0) - 0.8. The "guessing" process is to continue until the absolute value of (x - x0) is less than tol.
• For Reynolds>10^6, calculations are halted and the program generates a message about that using the error() function.

May I code it thus:

function PDROP = ex(rho, omega, length, diameter, Reynolds)
if (Reynolds<2320)
lambda = 64/Reynolds;
elseif (Reynolds>2320 && Reynolds<10^5)
lambda = 0.3164/(Reynolds^0.25);
elseif (Reynolds>10^5 && Reynolds<10^6)
tol = 0.01;
x0 = 1;
x = 2*log10(Reynolds) - 0.8;
while (abs(x - x0) > tol)
x0 = x;
x = 2*log10(Reynolds/x0) - 0.8;
end
lambda = x^(-2);
elseif (Reynolds>10^6)
error('Reynolds number must not be larger than 10^6');
end
PDROP = 0.5*rho*(omega^2)*(length/diameter)*lambda;?