How do you solve #log_10 200!#?

Let #log_10 200=x#

Then by definition, #10^x=200#

Now take the log on both sides and simplify using laws of logs to get

#xlog10=log200#

#thereforex=(log200)/(log10)=log200=2.30103#

You can either ask a computer to evaluate #200!# then take the log of it, or you can sum the logs:

#log_10 (200!) = sum_(n=1)^200 log_10(n)#

Either way you have a large number of calculations to be done. The only advantage I can see of summing the logs of each number is that you will not have to deal with numbers larger than #10^100#, which is a limiting factor for some pocket calculators.

Here are some computer generated figures:

#200! = 788,657,867,364,790,503,552,363,213,932,185,062,295,135,977,687,173,263,294,742, 533,244,359,449,963,403,342,920,304,284,011,984,623,904,177,212,138,919,638,830, 257,642,790,242,637,105,061,926,624,952,829,931,113,462,857,270,763,317,237,396, 988,943,922,445,621,451,664,240,254,033,291,864,131,227,428,294,853,277,524,242, 407,573,903,240,321,257,405,579,568,660,226,031,904,170,324,062,351,700,858,796, 178,922,222,789,623,703,897,374,720,000,000,000,000,000,000,000,000,000,000,000, 000,000,000,000,000#

and #log_10(200!) ~~ 374.89688864#