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I need to rewrite two expressions in terms of exponentials and simplify.

First expression: 2cosh(ln x) --> Answer should be x + 1/x

Second expression: ln(cosh x + sinh x) + ln(cosh x - sinh x) --> Do not know answer

A step by step would be helpful, I really need to learn this and am sinking at the moment. Thank you in advance.

bankey
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1 Answers1

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This might help, (by definition):

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

Replace x with $\ln x$ and simplify, note that $e^{\ln x} = x$

Use a similar idea for your second expression and note that:

$$ \sinh x = \frac{e^x - e^{-x}}{2}$$

John_dydx
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  • Would I have to distribute the 2 into the expanded form of coshx? I know I don't have to at this point because it would cancel out anyway but say if it was a 4coshx, then I would have to distribute the 4 into each e^x correct? Sorry for my lack of knowledge of Latex. And final question: when I get e^-lnx, how does that become 1/x? – bankey Sep 05 '15 at 23:37
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    Correct! By distributive law, you will have to multiply through by the constant. $e^{-\ln x} $ is the same as $\frac{1}{e^{\ln x}}$. Does this help? – John_dydx Sep 05 '15 at 23:40
  • @John It helps a lot John thank you so much :) Would I be able to contact you personally if I had anymore questions? It wouldn't hurt my feelings if you said no – bankey Sep 05 '15 at 23:48
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    @user268303, I'm more than happy to assist you, you're most welcome anytime. – John_dydx Sep 05 '15 at 23:49
  • $e^{-ln(x)}=(e^{ln(x)})^{-1})=x^{-1}$ – Paddling Ghost Sep 06 '15 at 00:02
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    Thank you for clearing that up for me Paddling Ghost :), and I got 2x for the second expression, looks right to me – bankey Sep 06 '15 at 00:04
  • @user268303, the second expression should hopefully give you zero. $\cosh x + \sinh x = e^x$, $\cosh x - \sinh x = e^{-x}$ which leaves you with $\ln e^x + \ln e^{-x}$. Use $\ln a^n = n \ln a$. – John_dydx Sep 06 '15 at 00:14
  • @John How does coshx + sinhx give e^x and coshx−sinhx=e^−x?? When I distributed ln to all the exponents it gave me (e^lnx + e^-lnx + e^lnx - e^-lnx)/2 + (e^lnx + e^-lnx - e^lnx - e^-lnx)/2 which, after combining like terms gave me (x + x)/2 which simplified to x. - Do you have skype? – bankey Sep 06 '15 at 00:42
  • @user268303, you've done it wrong. $\cosh x + \sinh x = \frac{e^x+ e^{-x}}{2} + \frac{e^x - e^{-x}}{2}$. The point is you need to simplify the bracket first before applying $\ln$. – John_dydx Sep 06 '15 at 00:46
  • So if I simplified the first bracket it would be (2e^x)/2 which would simplify into just e^x. And the other equation would be just e^-x. I understand thank you :)))) – bankey Sep 06 '15 at 00:53
  • @user268303, you're very welcome. – John_dydx Sep 06 '15 at 01:08