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What is the value of $cosh(ln3)$ ?

1 Answer

Nov 8, 2016

$cosh(ln3)=5/3$

Explanation:

$Coshx=1/2(e^x+e^-x) AAx inRR$

so

$cosh(ln3)=1/2(e^(ln3)+e^-ln3)$

$cosh(ln3)=1/2(e^(ln3)+e^ln(1/3))$

$cosh(ln3)=1/2(3+1/3)=1/2(10/3)$

$cosh(ln3)=5/3$

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