When does I_C be approximately equal to I_E
You probably already know that the emitter current \$I_E\$ is actually the sum of the collector current \$I_C\$ and base current \$I_B\$

\$I_E = I_C + I_B\$

Now if we know the current amplification \$\beta\$ we can know that:

\$I_B = I_C / \beta\$

Then the expression for \$I_E\$ becomes:

\$I_E = I_C + I_C/\beta = (1 +1/\beta) * I_C\$

In this expression you can see that influence of the base current \$I_B\$ becomes smaller as \$\beta\$ increases. So when \$\beta\$ is large we can accept some error and just ignore the base current \$I_B\$. Then we just assumne \$I_B = 0\$ so we can use:

\$I_E = I_C\$

For example:

if \$\beta\$ = 10, \$I_B\$ has a value of \$\frac {1}{10}I_C\$ so using \$I_E = I_C\$ would mean we make an error of about 10%

if \$\beta\$ = 100, \$I_B\$ has a value of \$\frac {1}{100}I_C\$ so using \$I_E = I_C\$ would mean we make an error of about 1%

if \$\beta\$ = 1000, \$I_B\$ has a value of \$\frac {1}{1000}I_C\$ so using \$I_E = I_C\$ would mean we make an error of about 0.1%

So it depends on how accurate you need your result to be, if you can accept a 1% error then using \$I_E = I_C\$ is fine as long as \$\beta\$ > 100.

In many practical circuits you would use resistors that have an accuracy of (much) less than 1% so there is no need to take the base current into account if \$\beta\$ > 100 as the error introduced by resistors is usually much larger.