Why does my code return 'Empty sym: 0-by-1'?

Currently working through chapter 12 in the third edition of MATLAB for Engineers by Holly Moore. I was trying to replicate example 12.1, which shows how to solve the reaction rate constants equation for Q: using symbolic variables and the solve equation. After a few tries, I copied the given code word for word but every time I run it, it returns 'Empty sym: 0-by-1' instead of the result the book shows, which is

ans =
-R*T*log(k/k0)

My input is:

syms k0 Q R T
k = k0*exp(-Q/(R*T)) == 0;
Q0 = solve(k,Q)

The book calls for:

X = sym('k = k0*exp(-Q/(R*T))')
solve(X,'Q')

I tried both, but neither returns what it's supposed to as far as I can tell

Answer 1

[Walter Roberson]

You have misframed k = k0 * exp(-Q/(R*T)) . Written as an equation equalling 0, it would be eqn = k - k0 * exp(-Q/(R*T)) == 0

syms k k0 Q R T
eqn = k  - k0 * exp(-Q/(R*T)) == 0

eqn = 

solve(eqn, Q)

ans = 

The way you had it was trying to solve k0*exp(-Q/(R*T)) == 0 for Q. That has the trivial solution k0 = 0. Otherwise divide by sides by k0 to get exp(-Q/(R*T)) == 0, so log(exp(-Q/(R*T)) == log(0) so -Q/(R*T) == -inf . That in turn has the solution "Q non-negative, R or T or both are 0", or the solution "Q is inf, R and T both positive or both negative" or "Q is -inf, R and T opposite signs". There might be additional solutions in complex values. solve() could potentially have returned a piecewise() of all of the possibilities.

[Walter Roberson]

The sym() version is old-style, not supported since roughly R2017b. The closest modern approach to that would be

X = str2sym('k = k0*exp(-Q/(R*T))')

X = 

solve(X,sym('Q'))

ans = 

●

[Elena]

Oh wow okay, good to know. Reframing it fixed it, thank you!