Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top


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.

share|cite|improve this question
$\ln(e^x) = x$ 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: $$ x-2=\sqrt{x}+4 $$

share|cite|improve this answer

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

share|cite|improve this answer


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

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

share|cite|improve this answer
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

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.