# How can I transfer the exponential relationship to linear?

The exponential relationship between y and x I got is

    model_1 <- nls(y ~ I(a * exp(-b * x)),
start = list(a = 1, b = 0), trace = TRUE)


and I got the parameter of a and b


Formula: y ~ I(a * exp(-b * x))

Parameters:
Estimate Std. Error  t value Pr(>|t|)
a  9.997e-01  2.664e-06 375275.8   <2e-16 ***
b -2.134e-04  4.298e-07   -496.5   <2e-16 ***
---

Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1

Residual standard error: 0.00136 on 626679 degrees of freedom

Number of iterations to convergence: 2
Achieved convergence tolerance: 3.387e-08


How can I transform y or x so that if I replot them, there will be a linear relationship between them?

In terms of the relationship alone

$$y = a \exp(-b x)$$

implies

$$\ln y = \ln a - b x$$

so that you could regress $$\ln y$$ on $$x$$ and accordingly expect that the coefficient of $$x$$ will be negative. Exponentiating the estimated intercept will give you an estimate of $$a$$ in the original equation.

But whether this is a good thing to do depends on the variability around the line (in that space). The assumptions behind this include the variation of $$\ln y$$ being about constant with $$x$$, which you can assess in various ways. The simplest and usually most fruitful are looking at a scatter plot with $$y$$ on logarithmic scale and looking at a plot of residual versus fitted for the regression. If the variability is approximately constant on the original scale, then indeed nonlinear least squares is indicated.

Detail: With your sample size, it may be that (say) a 1% random sample gives a scatter plot that will be as easy or easier to think about.