If $\sinh x-\cosh x=5$, find $\tanh x$
I have done till the following steps but dont know how to proceed further from solving this equation in Euler's form $$\left(\frac{e^x-e^{-x}}{2}\right)-\left(\frac{e^x+e^{-x}}{2}\right)=5$$ $$\frac{\not{e^x}-e^{-x}-\not{e^x}-e^{-x}}{2}=5$$ $$\frac{-\not{2}e^{-x}}{\not{2}}=5$$ $$-e^{-x}=5$$ $$\log(e^{-x})=\log(-5)$$ $$-x=\log(-5)$$ $$x=-\log(-5)$$
But, according to answer, I have it say $$x=\frac{-\log(25)}{2}$$
I don't know where I am going wrong.
The question: If $\sinh x-\cosh x=-5$, find $\tanh x$.
Note that using the identity $$\cosh^2x -\sinh^2x=1 ,$$ we have that $(\cosh x-\sinh x )(\cosh x+\sinh x)=1$ and so $$-(\sinh x-\cosh x)(\cosh x+\sinh x)=1 \Rightarrow -(-5)(\cosh x+\sinh x)=1$$ the we have that $$\cosh x+\sinh x=\frac{1}{5} ,$$ and adding to $\sinh x-\cosh x=-5$ we have $$2\sinh x=\frac{1}{5}-5 =-\frac{24}{5}\Rightarrow \sinh x=-\frac{12}{5} $$
Then, $\cosh x=\sinh x +5=-\frac{12}{5}+5=\frac{13}{5}$ and so $$\tanh x=\frac{\sinh x}{\cosh x}=-\frac{12}{13}. $$
First of all, change the signs to obtain:
$$\cosh(x) - \sinh(x) = -5$$
Now by the well known formula $e^{-x} = \cosh(x) - \sinh(x)$ you may rewrite
$$e^{-x} = -5$$ namely $$e^x = -\dfrac{1}{5}$$
and finally
$$x = \ln\left(-\dfrac{1}{5}\right) = i\pi - \ln(5)$$
At this point:
$$\tanh(x) = \tanh(i\pi - \ln(5)) = \dfrac{\tanh(i\pi) + \tanh(-\ln(5))}{1 + \tanh(i\pi)\tanh(-\ln(5))}$$
You know that $\tanh(i\pi) = 0$ thus the result is:
$$\tanh(x) = \tanh(-\ln(5)) = -\dfrac{12}{13}$$
