如何在R中的expoetial方程中获得x的值

Question

I have the following equation and I want to get the value of x.我有以下等式，我想得到 x 的值。

 (1/(1+exp(-(0.1348*x + 64.7027))))+ (x-70)=0

I try library(nleqslv), but I was unsuccessful to get x我尝试了 library(nleqslv)，但没有成功获得 x

Answer 1

Define a function f :定义一个函数f ：

f <- function(x) 1/(1+exp(-(0.1348*x + 64.7027))) + (x - 70)

In order to see where a root might fall, plot the function, trying several limits.为了查看根可能落在何处，绘制函数，尝试几个限制。

curve(f, from = 0, to = 100)

The one above has end points of opposite signs, so it's a job for uniroot .上面那个有相反符号的端点，所以这是uniroot的工作。

uniroot(f, interval = c(0, 100))
#$root
#[1] 69
#
#$f.root
#[1] 0
#
#$iter
#[1] 1
#
#$init.it
#[1] NA
#
#$estim.prec
#[1] 69

In order to have the value of the root, try any of the two ways below.为了获得根的值，请尝试以下两种方法中的任何一种。

uniroot(f, interval = c(0, 100))$root
#[1] 69

y <- uniroot(f, interval = c(0, 100))
y$root
#[1] 69