Question 26905
i have a feeling, from way back, that the method of solving this is a numerical one, ie trial and error.


x______xlogx
1 -->   0      TOO SMALL
2 -->   0.6020 TOO SMALL
3 -->   1.4314 TOO SMALL

jump and try
5 -->   3.49 TOO SMALL
6 -->   4.67 TOO SMALL
7 -->   5.91 TOO BIG

so the answer lies between 6 and 7 --> about in the middle, so lets try:
6.4 --> 5.16 TOO BIG
6.3 --> 5.04 TOO BIG
6.2 --> 4.91 TOO SMALL


so the answer lies between 6.2 and 6.3 --> closer to 6.3 so let's try:
6.28 --> 5.01 TOO BIG
6.27 --> 4.9989 TOO SMALL


so the answer lies between 6.27 and 6.28 --> very close to 6.27, so let's try:
6.271 --> 5.000099 TOO BIG


so the answer lies between 6.270 and 6.271 etc etc to however many decimal points you want. If you want an answer to quote, then find the mid point...eg quote 6.2705 as your answer or quote the nearest, in this case 6.271.


jon.