Question

can anyone prove where the fallacy lies here?
\[xlnx= lnx+lnx+.....x time\]diff. w.r.t x\[lnx+1=\frac{1}{x}+\frac{1}{x}+\frac{1}{x}+....x time\]\[lnx=0\]\[x=1\] but i didn't assume x to be 1

Answer 1

i know the answer you guys please try...

Answer 2

So we have

(x) (ln x) = (ln x) (ln x) ... (ln x) <- x many (ln x)
so we take d/dx [ (x) (ln x) = d/dx [ (ln x) (ln x) ... (ln x) ] <- x many (ln x)
and we obtain
(ln x) + 1 = 1/x + 1/x + ... + 1/x <- x many times
(ln x) + 1 = x ( 1/x )
ln x + 1 = 1
ln x = 0
exp (ln x) = exp (0)
x = exp (0)
so therefore x = 1

I don't see it

Answer 3

The obvious red flag is 1/x, what if x=0? But the preceding line has nothing but logs in it, and ln (0) doesn't make sense, so that's not even a possibility because the log function is restricted to natural numbers

Answer 4

one minute..how can write x(lnx)=(lnx)(lnx)(lnx)....x times???
it should be lnx +lnx +lnx..

Answer 5

Doesn't (ln x) + (ln x) + (ln x) + ... imply that it is infinitely repeating? Or no?

Answer 6

Why does it matter than you didn't assume x to be 1?

That shouldn't change the fact that it has to be 1?

Answer 7

xln(x)= ln(x)+ln(x)+ . . . + ln(x) (x times) is only defined for integers.

It is true that:
3ln(x)= ln(x)+ln(x)+ln(x)

But what if x = 3/2? x=sqrt(2)?

Thus this definition:
xln(x)=ln(x)+ln(x)+ . . . + ln(x) (x times) is not continuous, and taking the derivative of it doesn't make sense.

Answer 8

I agree with joemath. And to avoid that we can do it as follows:
\(n\ln{x}=\ln{x}+\ln{x}+...+\ln{x} \text{ (n times)}\). Now, we will differentiate with respect to x which is a continuous function defined for all \(x>0\):
\(\frac{n}{x}=\frac{1}{x}+\frac{1}{x}+...+\frac{1}{x} \text{ (n times}).\). You can see here that the relation is valid for all \(x>0\).

Answer 9

what joemath and Anwar have said is true..ues the fallacy lies there and one more thing is i have written \[xlnx=lnx+lnx+...xtime\] and again i am differentiating w.r.t x that is impossible.