One way to do that is to use the exponent rules for powers of the same base and for powers of powers. For example, I can rewrite $4^{27}$ as $(2^2)^{27}$, which is the same as $2^{54}$. Now I can cancel out some of the factors of 2 in the numerator and denominator. I get $2^{2001}\times 5^{1950}\div 4^{27} = 2^{2001-54}\times 5^{1950} = 2^{1947}\times 5^{1950}$. This is still a large expression, but it is easier to work with. I want to find out how many digits are in this value, but I don't want to calculate it exactly. That would take too long and be unnecessary. I recall that the number of digits in a positive integer is one more than the floor of its base 10 logarithm. That is, if $n$ is a positive integer, then $\text{digits}(n) = \lfloor \log_{10} n \rfloor + 1$, where $\lfloor x \rfloor$ is the largest integer less than or equal to $x$. So I can apply this formula to the expression $2^{1947}\times 5^{1950}$. I get $\text{digits}(2^{1947}\times 5^{1950}) = \lfloor \log_{10} (2^{1947}\times 5^{1950}) \rfloor + 1$. Now I can use another logarithm rule, which is that $\log_{10} (ab) = \log_{10} a + \log_{10} b$. This means I can split the logarithm of the product into the sum of the logarithms of the factors. I get $\text{digits}(2^{1947}\times 5^{1950}) = \lfloor \log_{10} 2^{1947} + \log_{10} 5^{1950} \rfloor + 1$. One more logarithm rule is that $\log_{10} b^a = a\log_{10} b$. This means I can bring down the exponents as coefficients of the logarithms. I get $\text{digits}(2^{1947}\times 5^{1950}) = \lfloor 1947\log_{10} 2 + 1950\log_{10} 5 \rfloor + 1$. Now I need to find the values of $\log_{10} 2$ and $\log_{10} 5$. I could use a calculator or a table of logarithms, but I also remember that $\log_{10} 2 \approx 0.3$ and $\log_{10} 5 \approx 0.7$. These are rough estimates, but they should be close enough for this purpose. I get $\text{digits}(2^{1947}\times 5^{1950}) = \lfloor 1947\times 0.3 + 1950\times 0.7 \rfloor + 1$. I can do this arithmetic in my head or on paper. I get $\text{digits}(2^{1947}\times 5^{1950}) = \lfloor 584.1 + 1365 \rfloor + 1$. I get $\text{digits}(2^{1947}\times 5^{1950}) = \lfloor 1949.1 \rfloor + 1$. I get $\text{digits}(2^{1947}\times 5^{1950}) = 1949 + 1$. I get $\text{digits}(2^{1947}\times 5^{1950}) = 1950$.


How many digits are in the value of the following expression: $2^{2001}\times 5^{1950}\div 4^{27}$?