Calculating bits of entropy of a passphrase with an equation for the bits of a random password is dumb. I'm calculating like log2((64k word dictionary size) ^ 5

Calculating bits of entropy of a passphrase with an equation for the bits of a random password is dumb. I'm calculating like log2((64k word dictionary size) ^ 5) = 80 bits. https://www.wolframalpha.com/input?i=log2%28%2864000%29+%5E+5%29 https://www.wolframalpha.com/input?i=log2%28%2864000%29+%5E+... I guarantee 5 words will be easier to remember than the 14 random characters required to get equal entropy from a random password from a dictionary of 58 characters. https://www.wolframalpha.com/input?i=solve+for+x+log2%28%2858%29+%5E+x%29+%3D+80 https://www.wolframalpha.com/input?i=solve+for+x+log2%28%285...