Least Common Denominator: $ \frac{\sqrt{x}}{x}+\frac{\ln\ x}{2\sqrt{x}} $

Question

$$ \frac{\sqrt{x}}{x}+\frac{\ln \ x}{2\sqrt{x}} $$

I have tried combining these two fractions; however, I keep getting stuck.

$$\frac{2\sqrt{x}}{2\sqrt{x}}\cdot\frac{\sqrt{x}}{x}+\frac{\ln\ x}{2\sqrt{x}}\cdot \frac{x}{x}$$

So that we get

$$\frac {2x}{2\sqrt{x}\cdot(x)}+\frac {x\ln\ x}{2\sqrt{x}\cdot(x)}$$

What am I doing wrong? The answer is $$ \frac{2+\ln\ x}{2\sqrt{x}}$$

Answer 1

Hint:

$\frac{\sqrt{2}}{2}=\frac{1}{\sqrt{2}}$

Sometimes one representation is more convenient for what you're trying to do than the other is.

Answer 2

The square root $\sqrt{x}$ seems to be the thing that bothers you: pretend it's not there! That is, write temporarily $\sqrt{x}=z$ so that $x=z^2$; then $$ \frac{\sqrt{x}}{x}+\frac{\ln x}{2\sqrt{x}} = \frac{z}{z^2}+\frac{\ln(z^2)}{2z} $$ Now, for avoiding misunderstandings, write temporarily $\ln(z^2)=\ln x=L$ so you have $$ \frac{z}{z^2}+\frac{L}{2z}=\frac{1}{z}+\frac{L}{2z}= \frac{2+L}{2z} $$ Substitute back to reinstate $x$: $$ \frac{2+L}{2z}=\frac{2+\ln x}{2\sqrt{x}} $$

This is longer, but safer. With some experience, you'll recognize more easily that $$ \frac{\sqrt{x}}{x}=\frac{1}{\sqrt{x}} $$ but the method of temporarily renaming objects to make them simpler works in several other cases.