How to apply one of the laws of logarithms to rearrange an equation?

Question

I want to rearrange the logarithm of a ratio shown in the paper here for figure 3 http://onlinelibrary.wiley.com/doi/10.1111/j.1469-7998.2006.00227.x/full

It is shown as log10(corneal diameter/ axial length) = -0.22

But I want to get it in terms of axial length.

Appendix 1 of this paper does just this http://www.jstor.org/stable/pdf/10.1086/670009.pdf and says:

axial length = corneal diameter * 10^-0.22

However, I've been criticised for using this and that it should be:

axial length=corneal Diameter/10^-0.22

So I turned to wolfram alpha and inputted:

log10(x/y)=-z, solve for y

which gave:

y = x * 10^0.22

equivalent to:

axial length = corneal diameter * 10^0.22 

So I have three different answers to what should be a trivial rearrangement. Much appreciated if you can tell me who's correct. Thanks.

Apologies if the links aren't accessible to all but I've tried to replicate everything in full here.

Answer 1

The last two answers are correct and hence are the same. Your equation is $$\mathrm{log}_{10}\frac{d}{L} = a$$ $$\frac{d}{L} = 10^{a}$$ $$L = \frac{d}{10^{a}} = \frac{d}{10^{-0.22}} = d\,10^{0.22}$$ We let $a=-0.22$, $d=$ corneal diameter, and $L=$ axial length.

Answer 2

Let's say that $a$ represents corneal diameter and $b$ represents axial length. Your aim is to find the value of $b$ from the equation $$ \log_{10} \left( \dfrac{a}{b} \right) = -0.22 \;. $$ Since $\log_{10} \left( \dfrac{a}{b} \right) = \log_{10} (a) - \log_{10} (b)$, we have $$ \begin{array}{rcl} \log_{10} (b) &=& \log_{10} (a) + 0.22 \\ &=& \log_{10} (a) + \log_{10} \left( 10^{0.22} \right) \\ &=& \log_{10} \left( a \cdot 10^{0.22} \right) \\ \end{array} \;. $$ Exponentiating on both sides, we obtain $$ b = a \cdot 10^{0.22} = \dfrac{a}{10^{-0.22}} \;. $$