Solving $e^{t/\!\ln(x)}=x$ for $x$

How do you solve the equation: $$e^{t/\!\ln(x)}=x$$ for $$x$$? Here $$t=1,2,3,\dotsc$$

This is what I did:

$$\ln(e^{t/\!\ln(x)})=\ln(x),$$

$$t/\!\ln(x)=\ln(x),$$

$$t=(\ln(x))^2,$$

$$x=e^\sqrt{t},e^{-\sqrt{t}}.$$

Is this correct? What kinds of numbers are the solutions?

• Yup, it is correct, but where $x$ belongs to?, I would think about the case when $x \in \Bbb{R}, x > 0$, but what about the equation when $x=1$?, it can't be right! – BAYMAX Sep 23 '18 at 4:57

You did perfect (apart from not noting that solutions must satisfy $$x>0$$ and $$x\ne1$$).

It might be easier if you set $$\ln x=y$$, so $$x=e^y$$. Then the equation becomes $$e^{t/y}=e^y$$ that yields $$\frac{t}{y}=y$$ and so $$y^2=t$$. Hence $$y=\sqrt{t}$$ or $$y=-\sqrt{t}$$ (assuming $$t>0$$, or the equation has no solution) and $$x=e^{\sqrt{t}}$$ or $$x=e^{-\sqrt{t}}$$.

• Here,x should be $x \ge 0~~and~~x\ne 1$ – Rakibul Islam Prince Sep 23 '18 at 9:45
• @RakibulIslamPrince $x>0$ and $x\ne1$; so long as $t>0$, the condition is satisfied for both solutions. – egreg Sep 23 '18 at 9:51
• Ohho.you didn't get my point...I am saying, $x\ge 0$. – Rakibul Islam Prince Sep 23 '18 at 9:54
• @RakibulIslamPrince No, I don't get it: in order that $\ln x$ exists it must be $x>0$. – egreg Sep 23 '18 at 10:00
• I agree,but,though $ln(x)$ is undefined at x=0,but the equation gives us a value of 1.look at it. – Rakibul Islam Prince Sep 23 '18 at 10:16

Yes,but as you are taking $$\ln$$ then you are assuming that x is positive and though it must have to be for the equation be valid.if we consider $$f(x)=e^{\frac{t}{\ln(x)}}-x$$ Then,the domain of this function. $$x\ge 0~~and~~x \ne 1$$. So,to be a solution your x value must have to be in this domain. You can also visualize it by it's graph.

And from the graph you can also verify your answer,that the curve is intersecting with x axis only on $$e^{\sqrt{t}}~~and~~e^{-\sqrt{t}}$$depending upon t.