This might be an easy problem for some, but I just can't wrap my head around this.
Prove through algebra, that any linear combination of the functions $e^x$ and $e^{-x}$ (for example: $f(x)=ae^x + be^{-x}$) can be written as $f(x)=d\cosh(x-x_0)$ given that a,b>0
On
Making
$$ a = e^{\lambda_1}\sigma(a)\\ b = e^{\lambda_2}\sigma(b) $$
with $\sigma(\cdot)$ the sign function, and making $\lambda = \frac{\lambda_1-\lambda_2}{2}, \lambda_0 = \frac{\lambda_1+\lambda_2}{2}$ we have
$$ a e^x+b e^{-x} = 2e^{\lambda_0}\left(\frac{\sigma(a)e^{x+\lambda}+\sigma(b)e^{-(x+\lambda)}}{2}\right) $$
Recall that $e^x=\cosh x+\sinh x$ and $e^{-x}=\cosh x-\sinh x$, so your function can be written $$ a(\cosh x+\sinh x)+b(\cosh x-\sinh x)= (a+b)\cosh x+(a-b)\sinh x $$ and we want to transform it into $$ k\cosh(x-x_0)=k\cosh x\cosh x_0-k\sinh x\sinh x_0 $$ so we need $$ \begin{cases} a+b=k\cosh x_0\\[4px] a-b=-k\sinh x_0 \end{cases} $$ Summing and subtracting, $$ \begin{cases} k(\cosh x_0-\sinh x_0)=2a \\[4px] k(\cosh x_0+\sinh x_0)=2b \end{cases} $$ Now we can multiply and recall that $\cosh^2x_0-\sinh^2x_0=1$, so we get $$ k^2=4ab $$ which gives $k=2\sqrt{ab}$ (because $k$ must be positive).
Finally, $$ \cosh x_0=\frac{a+b}{2\sqrt{ab}} $$ Note that a solution exists, because for positive $a$ and $b$, $a+b\ge2\sqrt{ab}$ (AM-GM inequality). The only thing to check is whether $x_0$ should be positive or negative. But we also have $$ \sinh x_0=\frac{b-a}{2\sqrt{ab}} $$ so we'll have $x_0>0$ if $b>a$, $x_0<0$ if $b<a$. If $b=a$ there's not much to do: $$ ae^x+ae^{-x}=2a\cosh x $$ so $x_0=0$.
Once you have transformed the function into a linear combination of $\cosh$ and $\sinh$, the procedure is not much different from the one with (trigonometric) cosine and sine. Suppose $$ f(x)=A\cos x+B\sin x $$ and we want to write it as $f(x)=k\cos(x-\varphi)$. The second expression translates into $k\cos x\cos\varphi+k\sin x\sin\varphi$, so we want $$ k\cos\varphi=A\qquad k\sin\varphi=B $$ Here we exploit $\cos^2\varphi+\sin^2\varphi=1$, so $$ k^2=A^2+B^2 $$ and we can choose $k=\sqrt{A^2+B^2}$ and we also have $$ \cos\varphi=\frac{A}{\sqrt{A^2+B^2}} \qquad \sin\varphi=\frac{B}{\sqrt{A^2+B^2}} $$ which determine $\varphi$.
Final note: $\cosh(u-v)=\cosh u\cosh v-\sinh u\sinh v$; indeed, the right hand side can be written as $$ \frac{e^{u}+e^{-u}}{2} \frac{e^{v}+e^{-v}}{2} - \frac{e^{u}-e^{-u}}{2} \frac{e^{v}-e^{-v}}{2} $$ Do the multiplications and simplify recalling $e^{x+y}=e^xe^y$.