I am not a mathematician, but I have a basic question about comparing numbers with exponents near $1$. I am evaluating how well a "model" works and I want to make a compelling case. For example, my model suggests
$$\frac{x_1}{x_2} = \sqrt{\frac{L_1}{L_2}}$$
Given $L_1$ and $L_2$, my model predicts the ratio $\frac{x_1}{x_2}$, that I will call $X_{ideal}$. This is what I will hopefully find when I take the quotient of the values I find for $x_1$ and $x_2$. So $X_{ideal}$ is the ideal, expected situation given those two values, $L_1$ and $L_2$.
But I measured $x_1$ and $x_2$, and calculated the actual ratio, which I'll call $X_{actual}$. $X_{actual}$ does not quite equal $X_{ideal}$, but the two numbers are close. I thought perhaps I could convince someone how close they were using exponents. I took the original model and just called $\frac{x_1}{x_2} = X_{ratio}$.
$$X_{ratio}^n = \sqrt{\frac{L_1}{L_2}}$$
Obviously $n$ is $1$ when $X_{ratio} = X_{ideal}$, because the model predicts $\sqrt{\frac{L_1}{L_2}}= X_{ideal}$ when $L_1$ and $L_2$ are known. Then I put in my $X_{actual}$ value though and wanted to see how close its exponent $m$ was to $1$: $$X_{ideal} = \sqrt{\frac{L_1}{L_2}} = X_{actual}^m$$
Turns out, it is really close! $m=0.9937$. Is that convincing? I am not sure why I did that, so is this even a good way to convince someone that my model predicted close to the actual values, or is it just unnecessarily complicated?
In fact, the question you address here is relevant from regression analysis. Basically, your assumed model is $$Y=X^m$$ which is nonlinear with respect to parameter $m$. The question is : is a value of $m$ close to $1$ significant ? There are statistical tests to check for that.
In any manner, remember that $100^{0.99}=95.5$, $1000^{0.99}=933.3$, $10000^{0.99}=9120.1$ and that shows significant deviations compared to the values obtained for $m=1$.
Among the classical tests, you have the standard deviation on parameters.
In any manner, for illustration, report on the same graph the experimental data as well as the line $Y=X$ and the curve $Y=X^m$. This could easily show the validity of your model.
For illustration purposes, I generated data using $X=100, 200,\cdots, 1000$, $m=0.99$ and the exact $Y$ values where changed using a random relative error between $-5$% and $+5$%. The nonlinear regression leads to $m=0.992815$ (estimated parameter error is $0.001321$) and the corresponding confidence interval is $\left( \begin{array}{cc} 0.989826 & 0.995804 \end{array} \right)$ which reveals that the model is very significant.