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I have two equations:

$a_1 = \sqrt{b_1}$ and $a_2 = \sqrt{b_2}$

I divde these

$\frac{a_1}{a_2} = \frac{\sqrt{b_1}}{\sqrt{b_2}}$

and solve for $a_1$

$a_1 = a_2 \frac{\sqrt{b_1}}{\sqrt{b_2}}$

Fine. Now I take the same initial equations, but logarithmise these

$Ln(a_1) = Ln(\sqrt{b_1})$ and $Ln(a_2) = Ln(\sqrt{b_2})$

Again divde these

$\frac{Ln(a_1)}{Ln(a_2)} = \frac{0.5 \, Ln(b_1)}{0.5 \, Ln(b_2)}$

and solve for $Ln(a_1)$

$Ln(a_1) = Ln(a_2) \frac{Ln(b_1)}{Ln(b_2)}$

Applying logarithm laws results in

$a_1 = a_2^{\frac{Ln(b_1)}{Ln(b_2)}}$

If this is correct, it is sort of confusing that $a_1$ can expressed in a linear, but also in an exponential form. Or is it simply that in the second case the initial logarithmisation is delogarithmised by the exponential function? Interestingly, also, the square root disappears in the logarithmised form.

Mockup Dungeon
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1 Answers1

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$$ a_2^{\ln(b_1)/\ln(b_2)}=e^{\frac{\ln(b_1)}{\ln(b_2)}\ln(a_2)} $$ Using $\displaystyle \ln(a_2)=\frac{\ln(b_2)}{2}$ you have

$$a_2^{\ln(b_1)/\ln(b_2)}=e^{\frac{\ln(b_1)}{2}}=\sqrt{b_1}=a_1$$

In fact, it is not surprising because you wrote $$ \frac{a_1}{a_2}=\sqrt{\frac{b_1}{b_2}} $$ So taking the log here gives you $$ \ln\left(\frac{a_1}{a_2}\right)=\frac{1}{2}\ln\left(\frac{b_1}{b_2}\right)=\frac{1}{2}\ln\left(b_1\right)-\frac{1}{2}\ln(b_2)=\ln(a_1)-\ln(a_2) $$

Atmos
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  • Thanks – I think I get the square root part. However, a main confusion is to understand how in one case I arrive at a linear relationship for $a_1$ and in the second case at an exponential relationship for $a_1$. So I am wondering this is because in the second case, my starting value actually is not $a_1$, but $Ln(a_1)$. Or, maybe more clear: If I input values for $a_2$, $b_1$ and $b_2$ in both $a_1$ relationshops, the result is different, and I'm not entirely sure why. – Mockup Dungeon Mar 07 '18 at 14:24
  • Yes, that's just because $ln(a^b)=bln(a)$ in fact a product is changed into a power. That's just it. – Atmos Mar 07 '18 at 15:15
  • Right. This is done for a physical process. I just can't see the justification for this. I'll keep searching. Thanks for the help. – Mockup Dungeon Mar 08 '18 at 21:32