Question

$xlnsqrt{x}=1$

Explanation:

Step1: Simplify the radical term

First, rewrite $\sqrt{x}$ as $x^{\frac{1}{2}}$. Then the logarithm becomes:
$$\ln x^{\frac{1}{2}} = \frac{1}{2}\ln x$$
The equation now is:
$$x \cdot \frac{1}{2}\ln x = 1$$

Step2: Rearrange into standard form

Multiply both sides by 2 to eliminate the fraction:
$$x\ln x = 2$$

Step3: Rewrite using exponential identity

Recall that $x\ln x = \ln(x^x)$. Substitute this into the equation:
$$\ln(x^x) = 2$$
Exponentiate both sides with base $e$ to cancel the logarithm:
$$x^x = e^2$$

Step4: Solve for $x$

Notice that if $x=e$, then $e^e
eq e^2$. Instead, use the Lambert W-function (the inverse of $f(z)=ze^z$). Rewrite $x\ln x=2$ by letting $t=\ln x$, so $x=e^t$:
$$e^t \cdot t = 2$$
This is of the form $t e^t = 2$, so $t=W(2)$, where $W$ is the Lambert W-function. Substitute back $t=\ln x$:
$$\ln x = W(2)$$
$$x = e^{W(2)}$$
Since $W(2)e^{W(2)}=2$, we can also express $x = \frac{2}{W(2)}$. For the real solution, $W(2)\approx0.8526$, so $x\approx e^{0.8526}\approx2.345$.

Answer:

The real solution is $\boldsymbol{x = e^{W(2)}}$ (or approximately $\boldsymbol{2.345}$), where $W$ denotes the Lambert W-function.