Find the value of hyperbolic $\tanh x$ function from the equation 
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.
 A: 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
$$\tanh x=\frac{\sinh x}{\cosh x}=-\frac{12}{13}. $$
A: 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:
$$\tanh(x) = \tanh(-\ln(5)) = -\dfrac{12}{13}$$
