##Pola Kebalikan Eksponensial

a <- 5
b <- 8
x <- rnorm(100, 50, 1.5)
error <- rnorm(100, 5, 5)

Persamaan Kebalikan Eksponensial

knitr::include_graphics("C:/Users/Hafly Akeyla Pari/Downloads/WhatsApp Image 2024-06-05 at 20.37.08_e19cec65.jpg")

y <- a*(exp(b/x))+error

plot(x,y)

Transformasikan kedua ruas di ln kan Y* = lnY x* = 1/x

## Warning in log(y): NaNs produced
##         y_tr       x_tr
## 1  1.5913852 0.01970099
## 2 -0.1196263 0.01948082
## 3  2.4327246 0.02005677
## 4  2.8924515 0.01981574
## 5  1.7465069 0.01974808
## 6  2.3951484 0.02130081
model_tr <- lm(y_tr~x_tr,dataframe)
summary(model_tr)
## 
## Call:
## lm(formula = y_tr ~ x_tr, data = dataframe)
## 
## Residuals:
##     Min      1Q  Median      3Q     Max 
## -4.8749 -0.1714  0.2190  0.3990  0.9075 
## 
## Coefficients:
##             Estimate Std. Error t value Pr(>|t|)
## (Intercept)    2.943      3.010   0.978    0.331
## x_tr         -36.167    149.844  -0.241    0.810
## 
## Residual standard error: 0.8748 on 96 degrees of freedom
##   (2 observations deleted due to missingness)
## Multiple R-squared:  0.0006065,  Adjusted R-squared:  -0.009804 
## F-statistic: 0.05826 on 1 and 96 DF,  p-value: 0.8098

Implikasi dari Transformasi kedua ruas, maka _0=ln() dan _1=

b0 <- model_tr$coefficients[[1]]
b1 <- model_tr$coefficients[[2]]
b0;b1
## [1] 2.942784
## [1] -36.16676

Mengecek apakah nilai hasil transformasi balik sudah sesuai dengan nilai yang asli sebab _0=ln(), maka =e^_0

a_duga <- exp(b0)
a_duga
## [1] 18.96858