diff --git a/contents/solutions/AI606.md b/contents/solutions/AI606.md
index d9f8c54..22039b2 100644
--- a/contents/solutions/AI606.md
+++ b/contents/solutions/AI606.md
@@ -16,6 +16,6 @@ Aus den Formeln für Leistung und Effektivspannung in den Hilfsmitteln lässt si
 
 $\hat{U} = U_{\mathrm{eff}} \cdot \sqrt{2}$ eingesetzt in $P = \frac{U^2}{R}$
 
-$P = \frac{(\frac{\hat{U}_{Sender}}{\sqrt{2}})^2}{R} = \frac{\hat{U}_{Sender}^2}{2R} = \frac{78,8^2}{100} = \qty{62,0944}{\watt}$
+$P = \frac{(\frac{\hat{U}_{\mathrm{Sender}}}{\sqrt{2}})^2}{R} = \frac{\hat{U}_{\mathrm{Sender}}^2}{2R} = \frac{78,8^2}{100} = \qty{62,0944}{\watt}$
 
 Die richtige Antwort ist daher "Zirka $\qty{60}{\watt}$".
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