How do u decide which solutions are valid to a hyperbolic equation

Question

The equation we were solving in class was 7coshx=9-2sinhx. We can rewrite this as 45 (coshx)^2 -126(coshx)+85=0. Hence we get coshx=5/3 and coshx=17/15. Now we get 4 possible solutions . x=ln(1/3) ,ln(3),ln(5/3) and x=ln(3/5) . Since arcoshx is a function if we define that x>=1 . Then we get the following solutions that x= ln(3) ,ln(5/3). However if you convert the equation 7coshx=9-2sinhx into exponential form we get 3(e^x)^2 -8(e^x)-3=0. So our solutions are now x= ln(1/3) and x=ln(5/3). Which contradicts my previous claim . Can someone show me where I went wrong .

Answer 1

Let's assume $x$ is a real number. Write $\sinh x$ and $\cosh x$ in terms of $e^x$ and $e^{-x}$. Substitute in your given equation to obtain a quadratic equation in $e^x$. Discard 0 or any negative solution, because $e^x$ is always positive. Take the natural logarithm to find$ x.$

Answer 2

HINT

Avoid log, exp ;

Square both sides and put

$$ \cosh^2 x = 1 + \sinh^2 x $$

Recognize a quadratic equation in ...?