-3

Given $$x = y^\frac{1}{ln y}$$ $$x = ?$$

Is x is the approximation of e ?

rome101
  • 9
  • 5
  • Welcome to MSE. It will be more likely that you will get an answer if you show us that you made an effort. – José Carlos Santos Jun 06 '18 at 12:40
  • 1
    Hint: $\frac 1{\ln y}=\log_y e$ at least assuming that $y$ is real and $>0$. Not sure where "roots" come into it. – lulu Jun 06 '18 at 12:42
  • Is the intent of the question to show that the iteration $y_{n+1}=y_n^{1/\ln(y_n)}$ tends to $e$ ? And what has the iteration do to with the fact that $e$ is transcendental ? – Peter Jun 06 '18 at 12:44
  • I mean e to the power of ln y is equal to y – rome101 Jun 06 '18 at 12:46
  • The given expression approximates $e$ well , if $y$ itself is an approximation of $e$ – Peter Jun 06 '18 at 12:48
  • 3
    $x=y^{1/ln y}$ taking logarithm of both side we get , $\ln x = (\frac{1}{\ln y})\ln y=1$ so $x= e^1=e$. – sirous Jun 06 '18 at 12:53
  • But if you compute first the exponent part which is 1/y and then raised y by the quotient. The decimal varies – rome101 Jun 06 '18 at 13:02
  • @rome101 Yeah, but in math, you do not use approximations, you use exact value – ray lin Jun 06 '18 at 13:04
  • @rome101 it's only a decimal approximation when you use the calculator or computer to approximate the value but even then, when you plug it into the calculator, the value should still return $2.718281828 ... $ which is a very good approximation of $e$ for most purposes – ray lin Jun 06 '18 at 13:06
  • @rome101 if you want an even more precise approximation of $e$, click this link: http://www-history.mcs.st-and.ac.uk/HistTopics/e_10000.html – ray lin Jun 06 '18 at 13:08

1 Answers1

2

Please note that $ \ln(1) = 0 $ and $ \ln(0)$ is undefined thus,

For $ \{ x | x \ne 0, x\ne 1 , x \in \mathbb{C} \} $,
$ x $ is exactly equal to $e$

Proof: $$ \because \ln(x) $$ is the inverse of $ e^x $ , $$\ln(e^x) = x $$ and $$ e^{\ln(x)} = x $$ $$ \therefore y = e^{ln(y)} $$

$$ y^{\frac{1}{\ln(y)}} = (e^{ln(y)}) ^ {\frac{1}{\ln(y)}} $$ $$ = e^{\ln(y) * \frac{1}{\ln(y)}} = e^{1} $$

Thus, $ x = y^{\frac{1}{\ln(y)}} = e $

ray lin
  • 329
  • How do you get that "since" sign? – Oscar Lanzi Jun 06 '18 at 13:05
  • 1
    @OscarLanzi you use the 2 dollar signs and put " \because" (without the quotes), you should check out this link for more latex functions: https://math.meta.stackexchange.com/questions/5020/mathjax-basic-tutorial-and-quick-reference – ray lin Jun 06 '18 at 13:09