Solve an equation with $e^{(x-2)}=e^{4}\cdot e^{\sqrt{x}}$

Question

$$e^{(x-2)}=e^{4}e^\sqrt{x}$$

I know that $x = 9$ and I can show the calculations like this:

$$e^{(x-2)} = e^{\sqrt{x}+4}$$

and now I need to get the $x$ to the right side but I dont know how.

Answer 1

Hint: Apply $\log$ to both sides of the equality. Rememeber that $(\forall y\in \Bbb R)\left(\log (e^y)=y)\right)$.

To solve an equation that looks like $ay+b\sqrt y+c=0$, introduce the variable change $w=\sqrt y$ to get $aw^2+bw+c=0.$

Answer 2

Hint: Since the function $f(x)=e^x$ is injective (or one-to-one), we know that $e^a=e^b \implies a=b$. So we may equate exponents to obtain: $$ x-2=\sqrt{x}+4 $$

Answer 3

Hints:

$$x-2=\sqrt x+4\stackrel{t:=\sqrt x}\implies t^2-t-6=0\implies (t-3)(t+2)=0\;\ldots$$

Note that it must be $\,x\ge 0\,$ .