# Hyperbolic Function & e

• May 3rd 2008, 01:17 PM
Simplicity
Hyperbolic Function & e
Q: Starting from the definition of $\mathrm{cosh}x$ in terms of $e^x$, find, in terms of natural logarithms, the value for $x$ for which $5 = 3 \mathrm{cosh}x$.

I did:
$5 = 3 \mathrm{cosh}x$
$5 = 3 \left( \frac{e^x + e^{-x}}{2} \right)$
$10 = 3e^x + 3e^{-x}$
$0 = 3e^x + e^{-x} - 10$
$0 = 3e^{2x} - 10e^x + 1$

However, that does not factorise. The answer is $\pm \ln 3$. Can someone do this question to see if they can get the answer. If possible, please show me your steps. Thanks in advance. (Smile)
• May 3rd 2008, 01:24 PM
o_O
Quote:

Originally Posted by Air
I did:
$5 = 3 \mathrm{cosh}x$
$5 = 3 \left( \frac{e^x + e^{-x}}{2} \right)$
$10 = 3e^x + 3e^{-x}$
$0 = 3e^x + {\color{red}3}e^{-x} - 10$

Where did the 3 go? ;)
• May 3rd 2008, 01:27 PM
Simplicity
^ Thanks!

Also, is $\ln \left( \frac{1}{3} \right) = - \ln (3)$?
• May 3rd 2008, 01:29 PM
o_O
Yep: $\ln \left(\frac{1}{3}\right) = \ln \left(3^{-1}\right) = -\ln 3$

where: $\ln (a)^{n} = n\ln a$
• May 3rd 2008, 01:33 PM
Simplicity
Another follow on question is: Find the minimum value of $(5 \mathrm{cosh}x + 3 \mathrm{sinh}x$.

I have no idea how to do this. Can someone help me on how to do this. Thanks.
• May 3rd 2008, 01:38 PM
o_O
If you recall what you've learned from derivatives, you know that your local minimum can be found by finding f'(x) = 0 and verifying with your first derivatives test. So:

$f(x) = 5\left(\frac{e^{x} + e^{-x}}{2}\right) + 3\left(\frac{e^{x} - e^{-x}}{2}\right)$
$f(x) = \frac{5e^{x} + 5e^{-x} + 3e^{x} - 3e^{-x}}{2}$

Simplify and find for what values f'(x) = 0.