Question below...   A password can include small alphabets and numbers. If two special characters are allowed, the increase in the level of safety would be equal to increasing the length of the password by approximately how many letters if their use was not allowed.

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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.

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