How can i calculate Total no. of digit in $2^{100}\cdot 5^{75}$ 
How can i calculate Total no. of digit in $2^{100}\cdot 5^{75}$

$\bf{My\; Try::}$ I have used $$\log_{10}(2) = 0.3010$$.
Now Total no. of digit in $$x^y = \lfloor \log_{10}x^y\rfloor +1$$
Now  $$\log_{10}(2^{100}\cdot 5^{75}) = 100\cdot \log_{10}(2)+75\log_{10}(5) = 100\cdot \log_{10}(2)+75-75\log_{10}(2)$$
So we get $$\log_{10}(2^{100}\cdot 5^{75})=30.10+75=105.10$$
So We get no. of Digit in $$2^{100}\cdot 5^{75} = \lfloor \log_{10}(2^{100}\cdot 5^{75})\rfloor +1 = 105+1 = 106$$
Can we solve it without using $\log\;,$ If yes then plz explain me, Thanks

I will note that your answer went from $100 \log_{10}(2) + 75 - 75\log_{10}(20)$ to $100 \log_{10}(2) + 75$.  The general form of argument is correct, but taking this mistake into account the answer is $\lfloor 25\log_{10}(2) + 75\rfloor+1 = \lfloor 75 + 7.525...\rfloor+1 = 83$.
Doing this without logarithms is fairly straightforward.  Factor out the terms $2^{75}5^{75} = 10^{75}$, reducing the problem to adding 75 to the number of digits of $2^{25}$.  For this I will note $2^{10} = 1024$, so from the crude estimate $1000 < 2^{10} < 1500$, we can obtain:
$$
10^7 < 1000^2(32) < (2^{10})(2^{10})(2^5) < 1500^2 (32) = 1000^2(1.5)^2(32) = 1000^2(72)< 10^8
$$
Thus, the number of digits of $2^{25}$ is 8, and the answer is 83.
