3 years ago

Numerical value of cosh(ln5)

Mathematics

1 answer:

strojnjashka [21]3 years ago

6 0

$\cosh x=\dfrac{e^x+e^{-x}}2$

So

$\cosh(\ln 5)=\dfrac{e^{\ln5}+e^{-\ln5}}2=\dfrac{5+\frac15}2=\dfrac{13}5$

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