Number Of Digits Corresponds to Logarithm?

A direct way to do these type of questions,

$\lfloor \log 6^{120} \rfloor +1= \lfloor 120\log 6 \rfloor +1\\ \lfloor 120( \log 2+\log 3) \rfloor+1 = \lfloor120(0.7781) \rfloor+1 \\ \lfloor 93.372 \rfloor+1 =\boxed{94}$

120 log 2 = 36.12

120 log 3 = 52.25

120(log 2 + log 3) = 93.37

120(log 6)=93.37

log 6^120 = 93.37

10^93.37 = 6^120

therefore there are 94 digits

$(log 2 ) + (log 3 ) = log(2\times3) = log(6) = 0.3010 + 0.4771 = 0.7781$

$6^{120} = 6^{100+10+10} = 6^{100}\times6^{10}\times6^{10}$

$0.7781^{100} + 2\times(0.7781^{10}) = 93.372 ≈ 94$

This can be done without the values given by knowing a couple of useful pure power coincidences.

Clearly $6^{120} = 2^{120} \cdot 3^{120}$ .

The first coincidence to use is $3^{12} \approx 2^{19}$ . Therefore $3^{120} \approx 2^{190}$ .

Using the exponent rules we now have $6^{120} \approx 2^{310}$ .

The second coincidence to use is $2^{10} \approx 10^3$ . Therefore $2^{310} \approx 10^{93}$ .

Knowing the second approximation slightly reduced the quantity, and that the error of the first approximation is smaller than the error of the second, we have confidence that $6^{120} > 10^{93}$ therefore the true quantity will have 94 digits.

A very general way to do these type of question is

If we want to find the number of digits in $a^{b}$

Simply we can do

[b (log a)] + 1

Or if b(log a) is in decimals just round it off to the next integer without adding 1