Let A, B be uniform algebras. Suppose that A0, B0 are subgroups of A−1, B−1 that contain exp A, exp B respectively. Let α be a non-zero complex number. Suppose that m, n are non-zero integers and d is the greatest common divisor of m and n. If T : A0 → B0 is a surjection with \documentclass[12pt]{minimal}
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\begin{document}$${\|T(f)^{m}T(g)^{n} - \alpha\|_{\infty} = \|f^{m}g^{n} - \alpha\|_{\infty}}$$\end{document} for all \documentclass[12pt]{minimal}
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\begin{document}$${f,g \in A_0}$$\end{document}, then there exists a real-algebra isomorphism \documentclass[12pt]{minimal}
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\begin{document}$${\tilde{T} : A \rightarrow B}$$\end{document} such that \documentclass[12pt]{minimal}
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\begin{document}$${\tilde{T}(f)^d = (T(f)/T(1))^d}$$\end{document} for every \documentclass[12pt]{minimal}
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\begin{document}$${f \in A_0}$$\end{document}. This result leads to the following assertion: Suppose that SA, SB are subsets of A, B that contain A−1, B−1 respectively. If m, n > 0 and a surjection T : SA → SB satisfies \documentclass[12pt]{minimal}
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\begin{document}$${\|T(f)^{m}T(g)^{n} - \alpha\|_{\infty} = \|f^{m}g^{n} - \alpha\|_{\infty}}$$\end{document} for all \documentclass[12pt]{minimal}
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\begin{document}$${f, g \in S_A}$$\end{document}, then there exists a real-algebra isomorphism \documentclass[12pt]{minimal}
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\begin{document}$${\tilde{T} : A \rightarrow B}$$\end{document} such that \documentclass[12pt]{minimal}
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\begin{document}$${\tilde{T}(f)^d = (T(f)/T(1))^d}$$\end{document} for every \documentclass[12pt]{minimal}
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\begin{document}$${f \in S_A}$$\end{document}. Note that in these results and elsewhere in this paper we do not assume that T(exp A) = exp B.
