= Solution
The Euclidean matrix <norm> is the induced <operator norm> $\|C\|_2=\sup_{\|x\|_2=1}\|Cx\|_2$. By the <spectral theorem>, a real <symmetric matrix> has $A=Q\Lambda Q^T$ with $Q$ orthogonal and $\Lambda=\operatorname{diag}(\lambda_1,\ldots,\lambda_m)$. Hence
$$
e^{tA}=Q\operatorname{diag}(e^{t\lambda_1},\ldots,e^{t\lambda_m})Q^T.
$$
Orthogonal transformations preserve the Euclidean <norm>, and the induced <norm> of a real diagonal matrix is the maximum absolute diagonal entry. Therefore, for $t\geq0$,
$$
\boxed{\|e^{tA}\|_2=\max_j e^{t\lambda_j}=e^{t\mu(A)}.}
$$
This gives the requested inequality with equality. If an exponent $\beta<\mu(A)$ were to work with prefactor one for every $t$, choose a unit <eigenvector> for $\mu(A)$. Its image has length $e^{t\mu(A)}>e^{t\beta}$ for every $t>0$, a contradiction. Thus \b[$\mu(A)$ is the smallest possible exponent].
For <symmetric matrices> the <spectral abscissa> equals the <Euclidean logarithmic norm>. Symmetry is essential to this particular exact <norm> formula; a general non-normal matrix can have transient growth not controlled with prefactor one by its spectral abscissa.
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