Solve $9\cosh(x) - 5\sinh(x) = 9$

Question

I am having trouble solving $$9\cosh(x)-5\sinh(x)=9$$

I have done these steps already: $$9\left(\frac{e^x+e^{-x}}{2}\right) - 5\left(\frac{e^x-e^{-x}}{2}\right)=9$$ $$\frac{9e^x +9e^{-x}-5e^x+5e^{-x}}{2} =9 $$ $$2e^x + 7e^{-x} = 9$$

I am stuck on where to go next. (Provided the steps i've taken before are correct)

Thanks

Answer 1

HINT: multiply your equation by $$e^x$$ and solve a quadratic equation the equation is given by $$2e^{2x}-9e^x+7=0$$ can you finish this?

Answer 2

Hint : multiply by $y=e^x$ and solve the equation with $y$

Answer 3

Hint: Your equation is: $$ \frac{2e^{2x}-9e^x+7}{e^x}=0 $$ solve for $e^x=y$

Answer 4

Take $e^{x}=p$.

Then your equation becomes $$2p+\frac{7}{p}=9$$ $$2p^2-9p+7=0$$ $$p=\frac{9\pm\sqrt{81-56}}{4}=\frac{9\pm5}{4}=1,\frac{14}{4}$$ $$x=\ln \left(\frac{9\pm5}{4}\right)=0,\ln \frac{14}{4}$$

Hope this helps you.