Solving $2\cosh 2x+10\sinh 2x=5$ in logarithmic form gives $x$ equal to:

1. Write $\cosh 2x=\tfrac12(e^{2x}+e^{-2x})$ and $\sinh 2x=\tfrac12(e^{2x}-e^{-2x})$. 2. Substituting gives $(e^{2x}+e^{-2x})+5(e^{2x}-e^{-2x})=5$. 3. Collecting terms: $6e^{2x}-4e^{-2x}=5$, and multiplying through by $e^{2x}$ gives a quadratic in $e^{2x}$. 4. Solving that quadratic and keeping the positive root gives $e^{2x}=\dfrac43$. 5. Taking logarithms, $2x=\ln\dfrac43$, so $x=\dfrac12\ln\dfrac43$. The factor $\tfrac12$ comes from the $2x$ in the argument. _Source: NECTA ACSEE 2022 Advanced Mathematics 142/1, Question 2: Hyperbolic Functions_