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I have a question which is: $71\sinh(x)+29\cosh(x)=5$ where $x\in R$

At degree level how would I make this as mathematically concise and accurate as possible? Or is this acceptable?

$71sinh(x)+29cosh(x)=5$

$71(\exp(x)-\exp(-x))+29(\exp(x)+\exp(-x)=10$

$100\exp(x)-42\exp(x) = 10$

$50\exp(2x)-5\exp(x)-24=0$

$\exp(x)={\frac {5\pm\sqrt{25+4*50*24}}{100}}$

$\exp(x)=0.7$

$x=\ln({\frac7 {10}})$

james15c
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    Every graduate student (and I would say most of the undergraduates) should be able to fill in the missing steps if you just present the final result. Another thing is if you are the student and should present this result to a professor as part of an exam (in that case depends on the exam and on the professor) – Manlio Nov 08 '17 at 16:06

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Why don't you post your solution so we can comment on it?

My assumption is that you use the addition or subtraction formula for hyperbolic sine or cosine together with the fact that $\cosh^2-\sinh^2=1$ and then use the formula for the inverse hyperbolic function.

In this kind of problem you can give lots of detail as I often do or just give the results requiring the reader to fill in those details. And actually the previous paragraph slightly rephrased could be considered a complete solution to the problem though you might want to add the two statements that (1) if $a^2-b^2=1$ and $a>b>0$ there is an $x$ such that $\cosh x = a$ and $\sinh x =b$ and (2) if $a>b>0$ there is a $c$ such that $(a/c)^2-(b/c)^2=1$.

marty cohen
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