Use identity:
\[
\cosh x = \frac{e^x + e^{-x}}{2},\quad \sinh x = \frac{e^x - e^{-x}}{2}
\]
Then:
\[
\cosh x - 5 \sinh x = \frac{e^x + e^{-x}}{2} - 5 \cdot \frac{e^x - e^{-x}}{2} = \frac{e^x + e^{-x} - 5e^x + 5e^{-x}}{2} = \frac{-4e^x + 6e^{-x}}{2}
\]
Set this equal to 5:
\[
\frac{-4e^x + 6e^{-x}}{2} = 5 \Rightarrow -4e^x + 6e^{-x} = 10
\]
Multiply by \(e^x\):
\[
-4e^{2x} + 6 = 10e^x \Rightarrow 4e^{2x} + 10e^x - 6 = 0
\Rightarrow \text{Solve to find } x = -\ln 2
\]