Help with solving a sinh x equation please

[nil101]

Please can you help me to solve \(\displaystyle \L
1.5 - 2e^x = 3\sinh x{\rm for real values of x}\)

This is what I've got so far.

\(\displaystyle \L
3\left( {\frac{{e^x + e^{ - x} }}{2}} \right) + 2e^x - 1.5 = 0\)

\(\displaystyle \L
3\left( {e^x + e^{ - x} } \right) + 4e^x - 3 = 0\)

\(\displaystyle \L
3e^x + 3e^{ - x} + 4e^x - 3 = 0\)

\(\displaystyle \L
7e^x + 3e^{ - x} - 3 = 0\)

I dont have confidence that I'm right up to this point because I think there should be a quadratic equation to solve to give the'real values of x'.

Would someone be able to put me straight?

Thanks

 

[royhaas]

You used the wrong formula for $\displaystyle sinh(x)$. You want $\displaystyle (e^{x}-e^{-x})/2$.

 

[soroban]

Hello, nil101!

As royhaas pointed out, you used the wrong definition . . .
$\displaystyle \;\;$I'll correct it.

$\displaystyle \text{Solve: } 1.5\,-\,2e^x\;=\;3\sinh x\,$$\displaystyle \text{ for real values of }x.$

\(\displaystyle \L 3\left(\frac{e^x\,-\,e^{-x}}{2}\right)\;=\;1.5\,-\,2e^x\)

\(\displaystyle \L3e^x\,-\,3x^{-x}\;=\;3\,-\,4e^x\)

\(\displaystyle \L 7e^x\,-\,3\,-\,3e^{-x}\;=\;0\)

Multiply by \(\displaystyle e^x:\;\;\L7(e^x)^2 \,- \,3e^x\,-\,3\:=\:0\)


We have a quadratic . . . let $\displaystyle u\,=\,e^x$

$\displaystyle \;\;$Then we have: \(\displaystyle \L\,7u^2\,-\,3u\,-\,3\;=\;0\)

\(\displaystyle \text{Quadratic Formula: }\L\:u\;=\;\frac{-(-3)\,\pm\,\sqrt{(-3)^2\,-\,4(6)(-3)}}{2(7)} \;= \;\frac{3\,\pm\,\sqrt{93}}{14}\)


Back-substitute . . .

We have: \(\displaystyle \L\,e^x \:=\:\frac{3\,+\,\sqrt{93}}{14}\)$\displaystyle \;\;\Rightarrow\;\;x\:=\:\ln(0.9033117911) \:=\:$- 0.101902157

$\displaystyle \;\;$and: \(\displaystyle \L\,e^x\:=\:\frac{3\,-\,\sqrt{93}}{4}\)$\displaystyle \;\;\Rightarrow\;\;x\:=\:\ln(-0.464546483)$ . . . not a real number!

 

[nil101]

Thank you very much for helping me out.
It's very much appreciated.