e^x*lnx = 1
I've been trying to solve that but haven't succeeded.
Anyone wanna give it a shot? :p
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e^x*lnx = 1
I've been trying to solve that but haven't succeeded.
Anyone wanna give it a shot? :p
So this is the same as
ln(x)*e^x = 1 right?
That means ln(x)*e^x - 1 = 0
Put that in my TI-84, and the answer comes out to be x = 1.310
I'm not sure how to do it by hand, I usually do hard equation solving with my calculator.
Well, my calculator is the best that IB allows, and they don't allow something where you can just enter the equation into it and it tells you the exact answers and how to get it. I have a *Solver* function, which must be equal to 0 in order for it to be used, and it only gives approximate answers.
And ln(x) is the log(base e)x. I know that ln(e^x)) is simply x. But when they are multiplied together, I don't think the same rules apply.
I was provided with a free TI-84 Plus just for doing IB. I wouldn't spend money on a calculator which either a) wasn't allowed by IB or b)could be upgraded when I got to uni.
I'm not sure yet what kind of calculator I'll be allowed in my engineering courses next year, so I wont be getting one until then.
But ontopic, my least favourite portion of the math curriculum was the logs and e^x, so I may have forgotten a rule about cancellations, but I don't remember there being one when they were multiplied.
e^lnx = x
But if you like Texas Instruments like I do, get a TI-89, it's pure awesomeness! :)
http://education.ti.com/educationpor...us_ti89ti.html
Indeed interesting equation.. I've failed to solve it :(.. Curious to see the answer!
is x ln x power or its (e^x)(lnX)
lol here is my poor attempt
http://img243.imageshack.us/img243/8444/attempt.png
line 8??? log x = 10^x ????
yeah e^x cancels out.
Besides my answer was obviously wrong, but a attempt wasn't bad :).
I mean... at line 8 you rewrite log x as 10^x.... which isn't even true (e.g. x =1 ==> 0 = 10)
Yeah, I did them while half a sleep lol.
I showed my teacher this, she couldn't do it. I will attempt it again.
Zomg... just use netwon's method :)
Exact answer:
http://mathbin.net/46986
Following the latter derivation, you can use the alternating series approximation theorem to find how ever many decimal places you would like!
:)
Oh, but I see 2 variables there. That's not an exact answer, nor an answer at all, just a different way to express the function.
And, before anyone dares to say it has no exact answer, here's a proof there is one:
ln(x) is continous and deriving ]0, infinity[
e^x is continuous and deriving ]-inf, inf[
-> ln(x)*e^x is continous and deriving ]0, infinity[
thus lnx*e^x = 1 has a solution (only one in this case, derivative is always positive)
Actually it is an answer ;)! Use the theorem I mentioned to work out an approximation, my answer is exact. There is no better way to represent it, nor way to calculate it.
In fact, the calculators you guys used probably used that infinite sum, and the theorem to calculate it. Modified for speed of course.
^ There's just the infinite sum, which is not an answer, but another way to express the function.
^ We used the answer to calculate the answer? You only posted a way to find an approximation
^ Sorry but I don't understand.
:rolleyes:
You see the difference between the series you presented as a solution to the function e^x*lnx and these?
The first one has 2 variables: n, which is given values, but then there's x which gets no values.
In the wikipedia article you posted there is only one variable in the function, n or k in some.
It's a function if it has variables that are not given any values.
It's a number if all variables get known values.