Cosh Algebra

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zzinfinity

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Nov 12, 2009
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The problem is,
Show that 5e^x+4e^x can be written in the form A*Cosh(x+c) and solve for the constants A and c.

I rewrote the Cosh to get

5e^x+4e^x=A/2*(e^(x+c)+e^-(x+c))

Then I did some pseudomath and broke it up term by term to get the system

5e^x=A/2*(e^(x+c))

4e^x=A/2 *e^-(x+c)

Then I did a fairly long substitution process to solve for A and c which took about two pages of work. I feel like there must be an easier way to do this. Is there some identity I'm missing or something? Thanks!!
 
zzinfinity said:
The problem is,
Show that 5e^x+4e^x can be written in the form A*Cosh(x+c) and solve for the constants A and c.

I rewrote the Cosh to get

5e^x+4e^x=A/2*(e^(x+c)+e^-(x+c))

Then I did some pseudomath and broke it up term by term to get the system

5e^x=A/2*(e^(x+c))

4e^x=A/2 *e^-(x+c)

Then I did a fairly long substitution process to solve for A and c which took about two pages of work. I feel like there must be an easier way to do this. Is there some identity I'm missing or something? Thanks!!
Click to expand...

5e^x+4e^x = 9 e^x ..... is that really the problem statement?
 
Yes, that appears to be the problem statement. I was skeptical at first as well and thought it was a

typo. But if one solves for A and c, it does work out to \(\displaystyle 9e^{x}\)


The way you done it is as good as any, zz
 
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