how to find x in the question Sinh(x-3)=1 even if i expand it into e terms i cant seem to get it

How to find x in the question Sinh(x-3)=1 even if i expand it into e terms i cant seem to get it

I.

$$\small{\text{ $ \begin{array}{rcl} \sinh{(x-3)} &=& 1 \quad | \quad \sinh^{-1} \\ x-3 &=& \sinh^{-1}(1) \\ x &=& 3 + \sinh^{-1}(1) \\ x &=& 3 + 0.88137358702\\ x &=& 3.88137358702\\ \end{array} $ }}$$

II.

$$\boxed{ \small{\text{ $ \sinh(x)=\frac{1}{2}\cdot \left( e^x - e^{-x}\right) \qquad \sinh(x-3)=\frac{1}{2}\cdot \left( e^{x-3} - e^{-(x-3)}\right) $ }}}$$

$$\small{\text{$ \begin{array}{rcl} \sinh(x-3) &=& 1\\ \frac{1}{2}\cdot \left( e^{x-3} - e^{-(x-3)}\right) &=& 1\\ e^{x-3} - e^{-(x-3)} &=& 2 \\ e^{x-3} - \frac{1}{e^{x-3}} &=& 2 \quad | \quad u = e^{x-3}\\ u-\frac{1}{u} &=& 2 \quad | \quad \cdot u\\ u^2-1 &=& 2\cdot u \\ u^2 -2\cdot u - 1 &=& 0 \\ u_{1,2} &=& \frac{ 2\pm \sqrt{4-4\cdot(-1) } }{2}\\ u_{1,2} &=& \frac{ 2\pm \sqrt{2\cdot4 } }{2}\\ u_{1,2} &=& \frac{ 2\pm 2\cdot\sqrt{ 2 } }{2}\\ u_{1,2} &=& 1\pm \sqrt{ 2 }\\ \end{array} $}}$$

$$\small{\text{$ \begin{array}{rcll} u &=& e^{x-3} \quad & | \quad \ln\\ \ln(u) &=& (x-3) \cdot \ln(e) \quad & | \quad \ln(e) = 1 \\ \ln(u) &=& x-3 \\ \boxed{x = 3 + \ln(u)} \end{array} $ }}$$

$$\small{\text{$ \begin{array}{rcl|rcl} u_1 &=& 1+\sqrt{2} \quad & \quad u_2 &=& 1-\sqrt{2}\\ x_1 &=& 3 + \ln(u_1) \quad & \quad x_2 &=&3 + \ln(u_2) \\ x_1 &=& 3 + \ln(1+\sqrt(2)) \quad & \quad x_2 &=&3 + \ln(\underbrace{1-\sqrt(2)}_{<0\text{ no solution!}}) \\ x &=& 3 + \ln(1+\sqrt(2)) \\ x &=& 3 + \ln( 2.41421356237 ) \\ x &=& 3 + 0.88137358702\\ x &=& 3.88137358702\\ \end{array} $ }}$$