Find the numerical value of each expression.
(a) $ \cosh (\ln 5) $
(b) $ \cosh 5 $
in this problem, we want to determine what is the hyperbolic co sign Of the natural log of five. And what is the hyperbolic co sign of five. The hyperbolic cosine function is to find like this hyperbolic cosine of X is equal to E. To the X. Power plus E. To the negative X power all divided by two. So if we're going to find a hyperbolic coast sign of the log of five, we're simply going to substitute the log of five in everywhere you see X. So if the hyperbolic co sign of X equals E. T. D. X plus each and negative X over two. Then the hyperbolic co sign of the natural log of five. Okay, we're going to substitute lage five in for X. So we're going to have E. T. D. X. Which will be E. To the log of five plus E. To the negative X. So E to the negative log of five. All divided by two. Now, each of the negative log of five. When you have a negative exponent, that can be rewritten as a fraction where one is in the numerator and uh Then E to the positive Lage five would be in the denominator. So just like X to the -2 equals one over excited positive too. E. To the negative log of five equals one over E. Uh to the positive block of five. Now a lot of this is gonna simplify pretty nicely. Even though right now it might not look like it. E. And the natural log rhythm or inverse functions of each other. Uh so E to the log of five is simply five plus. Now we have one over E to the log of five. Once again E. And the natural algorithm or in verses of each other. So E to the natural log of five is simply just five. And so this green expression becomes 1/5. So we have five plus 1/5. All over. True. Well five plus 1/5 is five and 1/5 And we're dividing it by two. Let's change this up here to a an improper fraction. So this would be 26/5 26 5th divided by two is the same thing as 26 50 times one half. So we get 26/10 or 13 5th. So there is a the numerical answer for the hyperbolic co sign of the natural log of five. Okay, give us a little space to work here. Now we have to determine what is the hyperbolic co sign of five. Hyperbolic cosine of X. Is each of the X plus each negative X over two. So the hyperbolic co sign of five will simply be E to the fifth plus E. To the negative five. All divided by two. We want numerical answer. So we're simply going to put this into the calculator, putting each of the 50 plus each of negative fifth. All divided by two. In the calculator we get 74 Point to one. If we round to the nearest 100