a=log(sqrt(10)) = 1/2
b=log(sqrt(10)+1) =
c=log(sqrt(10)+2)
then
$$WRONG\qquad b^2=log(\sqrt{10}+1)^2=2log(\sqrt{10}+1)\qquad WRONG$$
As Alan has pointed out -
this previous answer is incorrect, question is not of the from log(x2) it is (log x)2 so it should be
$$b^2=[log(\sqrt{10}+1)]^2 \approx 0.38357$$
$$ac=\frac{log(\sqrt{10}+2)}{2}$$
Careful with b2!
$${\mathtt{b}} = {log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right) \Rightarrow {\mathtt{b}} = {\mathtt{0.619\: \!331\: \!048\: \!066\: \!094\: \!5}}$$
$${\mathtt{bsquared}} = {{log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}^{\,{\mathtt{2}}} \Rightarrow {\mathtt{bsquared}} = {\mathtt{0.383\: \!570\: \!947\: \!098\: \!647\: \!1}}$$
$${\mathtt{2}}{\mathtt{\,\times\,}}{log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right) = {\mathtt{1.238\: \!662\: \!096\: \!132\: \!189}}$$
Thank you Alan, i appreciate you pointing out this error.
I am surprised that you didn't use brackets in your answer because without brackets the meaning is not clear.
(log(sqrt(10)+1))^2=
$${{log}_{10}\left({\sqrt{{\mathtt{10}}}}{\mathtt{\,\small\textbf+\,}}{\mathtt{1}}\right)}^{\,{\mathtt{2}}} = {\mathtt{0.383\: \!570\: \!947\: \!098\: \!647\: \!1}}$$
So it did - another wrinkle for Andre to iron out!
Sorry Alan.
PS I have already sent Andre an email telling him of this probem. I am assuming that he puts calculator errors near the top of his priority list!