[Math] Finding the Function to a log-log plot

functionslogarithms

My maths is not very good, so please bare with me if the answer is obvious. I have a log-log plot below, which I have generated in R.

enter image description here

I now need to find what i believe to be called the inverse to the function of the plot, resulting in me obtaining the following equation:

$y = constant . x^m$

I understand that $m$ will be the gradient of the line (0.71), but how do I calculate the constant?

Best Answer

CORRECTION (the constant factor depends of the $\log$ basis considered!) :

If $\log$ means $\log_{10}$ on both axis then

you got $\ \log_{10}(y)=5.3+0.71\log_{10}(x)\ $ (since $\ \log_{10}\bigl(x^{0.71}\bigr)=0.71\;\log_{10}(x)$).

Take the 'antilog' $10^x$ to obtain $\ y\approx 10^{5.3}\ x^{0.71}$ or : $$\boxed{y\approx 200000\ x^{0.71}}$$ (because $\log_{10}(2)\approx 0.3$ and with $2$ digits precision)

else if the 'natural logarithm' i.e. $\ln$ was taken on both sides then :

You got $\ \ln(y)=5.3+0.71\ln(x)$.

take the exponential of this to obtain $\ y\approx e^{5.3}\ x^{0.71}$ or : $$\boxed{y\approx 200\ x^{0.71}}$$

(because $\,e^{a+b}=e^a e^b\,$ and $\,x^c= e^{c\ln(x)}$)

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[Math] Deriving a power equation from a log-log line equation

Supposing that both axis use b-basis logarithms we have : $$\log_b(y)=A+B\,\log_b(x)$$ using $\ b^{\log_b(y)}=y,\ b^{c+d}=b^c\,b^d\ $ and $\ B\,\log_b(x)=\log_b(x^B)\ $ we get : $$y=b^A\, x^B$$ So that your $\,y=e^A\,x^B$ equality was right for the natural logarithm $\,\ln=\log_e\,$ while the antilog expression ($y=10^{5.26}\,x^{0.7089}$) was the correct choice for the $10$-basis logarithm $\log_{10}$.

[Math] Finding value of (y) of logarithmic equation given (x)

Say you're trying to solve for $\theta$ (i.e., $x_2$) in $ r = a e^{b \theta}$ and are unable to find an analytical solution. You can use numerical root-finding. In Matlab and Octave fzero (documentation) solves for roots/zeros of simple univariate functions. The first step is to re-write your equation as $ 0 = a e^{b \theta}-r$. Then, with th0 as an initial guess, you can solve for a root via:

a = 1;
b = -5.7;
r = 0.5;
th0 = 1;
f = @(th)a*exp(b*th)-r;
[th_sol,fval,exitflag] = fzero(f,th0)

This returns 0.121604768519289 for th_sol. In this case, this is equivalent to the analytical solution $\theta = \text{log}(r/a)/b$.

Of course if you're solving for $y_2$ (i.e., $r$), you don't need to do anything special. Just evaluate:

a = 1;
b = -5.7;
th = 0.5;
r = a*exp(b*th)

Best Answer

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