Question below... A password can include small alphabets and numbers. If two special characters are allowed, the increase in the level of safety...

If the password can include small alphabets and numbers,
there are a total of 26 + 10 = 36 choices.

If the password
can have have N characters, the number of combinations possible is
36^N

When two extra special characters are allowed and the
length of the password remains the same, the number of combinations possible is
38^N.

The safety that a password provides can be taken as
the total number of combinations possible. The higher the number of combinations
possible, the higher is the safety it provides.

So we have
to determine by how much N has to increase with the number of options equal to 36 to
equal 38^N.

36^(N + x) =
38^N

take the log of both the
sides

=> log 36^(N + x) = log
38^N

=> (N + x) * log 36 = N * log
38

=> (N + x) / N = log 38 / log
36

=> N / N + x/ N =
1.015

=> 1 + x/N =
1.015

=> x/N =
0.015

=> x =
0.015*N

The increase in the number of characters is
dependent on the initial number of characters that made up the password and is equal to
approximately 0.015*N

For a password with an
initial length of N characters the increase in the length of the password to equal the
safety due to the increase in the number of options by 2 is approximately 0.015*N.