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You are not logged in. #1 20051214 09:57:29
logarithims, remind me again?I had understood that 7 ln 7 = ln 6^7 And that ln 6  ln 2 = ln 6/2 (or 3) A logarithm is just a misspelled algorithm. #2 20051214 11:12:13
Re: logarithims, remind me again?You need to use the powercoefficient rule first. Why did the vector cross the road? It wanted to be normal. #3 20051214 12:06:04
Re: logarithims, remind me again?Huh? I assumed 7 ln 2 was the correct answer because that was the answer in my book! Hang on let me check it again. A logarithm is just a misspelled algorithm. #4 20051214 12:14:42
Re: logarithims, remind me again?Heres the first problem: Last edited by mikau (20051214 12:18:10) A logarithm is just a misspelled algorithm. #7 20051214 19:28:22
Re: logarithims, remind me again?hi yaz mikau!!! #8 20051214 20:18:24
Re: logarithims, remind me again?hooray!!! i solved it!!! #9 20051215 05:54:59
Re: logarithims, remind me again?Aww.. dang you got here before me. I solved it this morning over a cup of coffee. (relaxation defined) Yeah the key is to multiply the equation by e^x and write it in the form of a quadratic, au^2 + bu + c where u = e^x. The solution to the quadratic will be 7 + sqrt(50), it can't be 7  sqrt (50) because that would be a negative number and e^x can never be negative. Thus 7 + sqrt(50) is the only solution. A logarithm is just a misspelled algorithm. #10 20051215 06:01:07
Re: logarithims, remind me again?But I'm not sure I see the mistake I made which you pointed out. A logarithm is just a misspelled algorithm. 