Karnage Posted October 20, 2005 Posted October 20, 2005 Here's problem I need help on. It relies mainly on one's intuition and intelligence. Can anone hlep me out and give a full, good explanation? Every number can be represented in the standard form such as \frac{100}{3} = 3.33 * 10^1. In this form the number x = a * 10^N, where 1<a<10, while N is an integer exponent. Find the standard form for x = 1234^1234. Include only three leading decimals in your answer, as in the previous example.
Karnage Posted October 20, 2005 Author Posted October 20, 2005 x = 1234^1234 heehee sry im kinda new to latex
Karnage Posted October 20, 2005 Author Posted October 20, 2005 rrrrrrrr its suposed to be x = 1234^1234 (sumthin's up with the coding)
ecoli Posted October 20, 2005 Posted October 20, 2005 did you mean [imath] 1234^{1234} ? [/imath] edit: you need a { and } around the exponent, or it'll only read the first number.
Karnage Posted October 20, 2005 Author Posted October 20, 2005 x = 1234 ^ 1234 express that in the terms stated in the problem above.
ecoli Posted October 20, 2005 Posted October 20, 2005 I think you're talking about scientific notation, correct? you want [imath]1234^{1234} [/imath] in sci not?
Karnage Posted October 20, 2005 Author Posted October 20, 2005 yes i want it in the form of notation as shown in the previous example
ecoli Posted October 20, 2005 Posted October 20, 2005 I think it might be [imath] 1.234 * 10^{3702} [/imath] but don't take my word for it.
cosine Posted October 20, 2005 Posted October 20, 2005 Karnage said: Do you have an explanation? 1234^{1234}=10^{x} 1234(\log(1234))=x 1234(\log(1234)) = 3814.6829070663730305378590465079... 1234^{1234} = (10^{3814})(10^{.6829070663730305378590465079...}) 10^{.6829070663730305378590465079...} = 4.8184467781382543701667186035719... 1234^{1234} = 4.8184467781382543701667186035719... * 10^{3814} Sorry about the large amount of digits, I'm sort of paranoid about inaccurate decimal representations, but I was too tired to figure out a probably obvious exact form. But anyway, hope this helps.
Karnage Posted October 20, 2005 Author Posted October 20, 2005 Yea I was thinking the same method and got the same answer. I was hoping for a more systematic and exact way of figuring it out though. But thnx anyway
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