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

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

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.

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$\ln(e^x) = x$ when $x > 0$, so $e^{(x-2)} = e^{\sqrt{x}+4} \Rightarrow x-2 = \sqrt{x}+4$ – kba Jun 22 '13 at 18:15
@kba As Git Gud pointed out, the identity $\ln(e^x)=x$ holds true even if $x\le0$. – Adriano Jun 23 '13 at 0:05

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: $$x-2=\sqrt{x}+4$$

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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 $aw^2+bw+c=0.$

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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 $\,x\ge 0\,$ .

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Also, since we have $x-6=\sqrt{x}$ and the right hand side of this equation is (by the definition of a square root) guaranteed to be nonnegative, we know that the left hand side must also be nonnegative so that $x-6\ge0 \iff x\ge6$. – Adriano Jun 22 '13 at 11:20
$x^2-13x+36=0$ has the solution $+-i\sqrt{23}$ – Mikael Jensen Jun 22 '13 at 12:22
@MikaelJensen, two things (1) What does your comment have to do with this question/answer/comment?, and (2) The quadratic you wrote down has the solutions $$\frac{13\pm\sqrt{169-144}}2=\frac{13\pm 5}2=\begin{cases}9\\{}\\4\end{cases}$$ – DonAntonio Jun 22 '13 at 12:36
I thought the equation was equal to x-6=sqrt{x} – Mikael Jensen Jun 23 '13 at 20:16
I thought the equation was equal to sqrt{x} – Mikael Jensen Jun 23 '13 at 20:19