(B) We know that $\cosh(x) = \frac{e^x + e^{-x}}{2}$.Substituting $x = \log 4$,we get $\cosh(\log 4) = \frac{e^{\log 4} + e^{-\log 4}}{2}$.Since $e^{\

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(B) We know that $\cosh(x) = \frac{e^x + e^{-x}}{2}$.Substituting $x = \log 4$,we get $\cosh(\log 4) = \frac{e^{\log 4} + e^{-\log 4}}{2}$.Since $e^{\log 4} = 4$ and $e^{-\log 4} = e^{\log(4^{-1})} = \frac{1}{4}$,$\cosh(\log 4) = \frac{4 + \frac{1}{4}}{2} = \frac{\frac{17}{4}}{2} = \frac{17}{8}$.