A problem with sinh where I keep getting 3 different answers

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SUMMARY

The discussion centers on solving the equation sinh(x) = π using the definition of the hyperbolic sine function, sinh(x) = (e^x - e^(-x))/2. The user initially attempted to isolate x by manipulating the equation incorrectly, leading to confusion with logarithmic properties. The correct approach involves substituting u = e^x, transforming the equation into a quadratic form, and then solving for u before applying the logarithm. This method clarifies the discrepancy between the user's calculations and results from calculators and WolframAlpha.

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Homework Statement


This isn't a homework problem, just a thought problem. I want to find a solution to the equation sinh(x)=pi.


Homework Equations


sinh(x)=(e^x-e^-x)/2


The Attempt at a Solution


Here is a picture of the work that I did, pretty much, I set the equation for sinh equal to pi, and I found (ln(2pi)/2). Thats all one part, not the natural log of 2pi/2, that's just stupid. Any way, calculators gave my a number along the lines of 1.05 or something like that, and when I had wolframalpha solve sinh(x)=pi, they gave me a number with i and some square roots that I don't know where they came from. I just want an explanation of what's going on. When I plugged (ln(2pi)/2) into the equation for sinh, I got pi, so I showed that this give pi, but wolframalpha gives another answer, and calculators say I'm wrong. What did I do wrong? Sorry about the pictures, but I don't know LaTex.

photo-2.jpg

photo2.jpg
 
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The first thing you did wrong was saying ln(2pi+e^(-x))=ln(2pi)+ln(e^(-x)). That's completely wrong. ln(a+b) is not ln(a)+ln(b). Take your original equation and substitute u=e^x, so 1/u=e^(-x). Now it's quadratic in u. Solve for u first. Then take the log.
 
Oh, I made the mistake, the property is ln(ab)=ln(a)+ln(b), fail, alright, thanks.
 

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