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Numerical value of cosh(ln5)
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4 votes
Numerical value of cosh(ln5)

User Wei
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1 Answer

4 votes

\cosh x=\frac{e^x+e^(-x)}2

So


\cosh(\ln 5)=\frac{e^(\ln5)+e^(-\ln5)}2=\frac{5+\frac15}2=\frac{13}5
User Dan Simon
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