Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ • π ƒ ¹ ² ³ °
 

You are not logged in. #4 20051222 08:48:44
Re: Is there any way to solve this algebraically?Hmm. I don't really know what I'm doing here, but I'll give it a go. Why did the vector cross the road? It wanted to be normal. #5 20051222 10:02:50
Re: Is there any way to solve this algebraically?What about the inverses of the functions? I don't know what to do next. If you combine mathsypersons stuff with mine, you get x1 = x/2, but I don't know if that's legal after I took the inverse. Last edited by John E. Franklin (20051222 10:06:27) igloo myrtilles fourmis #7 20051222 22:09:14
Re: Is there any way to solve this algebraically?Let lg2 represent logarithm to the base 2. Character is who you are when no one is looking. #10 20051223 07:13:55
Re: Is there any way to solve this algebraically?at the end of post#5 I had x1 = x/2, but is this allowed since igloo myrtilles fourmis #12 20051229 08:18:09
Re: Is there any way to solve this algebraically?We have two answers: IPBLE: Increasing Performance By Lowering Expectations. #13 20051229 08:27:47
Re: Is there any way to solve this algebraically?Geometric proof: Last edited by krassi_holmz (20051229 08:53:50) IPBLE: Increasing Performance By Lowering Expectations. #14 20051229 10:25:46
Re: Is there any way to solve this algebraically?And we'll proof that there doesn't exist line that intersects with 2^i more than 2 times. That's because IPBLE: Increasing Performance By Lowering Expectations. #15 20051229 10:34:39
Re: Is there any way to solve this algebraically?Condition for existing line that divides grafhic of f(x) more than 2 times is that is nessesery to exist at least one inflex point. IPBLE: Increasing Performance By Lowering Expectations. #16 20051229 10:38:11
Re: Is there any way to solve this algebraically?Could somebody proof the upper using analisis? IPBLE: Increasing Performance By Lowering Expectations. 