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anyway to solve this? if so what level of math is needed?
As micromass said, you can't really solve that type of equation algebraically. In some simple case you might just "spot" a solution by inspection (eg 2^x = x^2; x=2 or x=4).anyway to solve this? if so what level of math is needed?
No, there's still no algebraic solution.what if you logged both sides using log with 2 power base?
you get x = 5 lnx
is that easier to solve?
all of them....?BTW. How many real solutions are you expecting to get for that equation?
2 if you use calculus it seems atleast according to wolfram. been looking into the lambert w function seems waay out of my league for now. seems there are no simple answer with this. but it sure looks simple.Can you deduce how many there are?
[itex]x=\cos(x)[/itex] might also look simple, but it too cannot be solved algebraically. They're called transcendental equations.2 if you use calculus it seems atleast according to wolfram. been looking into the lambert w function seems waay out of my league for now. seems there are no simple answer with this. but it sure looks simple.
Yes that is correct, there are two real solutions. For some fairly small positive value of "x", x^5 will exceed 2^x, but it is important to understand that for large "x" that 2^x will eventually exceed x^5 (or any power of x for that matter).2 if you use calculus it seems atleast according to wolfram. been looking into the lambert w function seems waay out of my league for now. seems there are no simple answer with this. but it sure looks simple.