Solving equation containing natural logarithm's and roots.

[retlig]

Hello!

I´m at my wits end with this problem. I´ve tried solving it like 10 times over multiple days. Can anyone nudge me in the right direction because I´m obviously missing some crucial knowledge to solve it.

$$\sqrt{\ln x}=\ln\sqrt{x}$$
Cheers!

 

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Is this schoolwork?

Either way, please show one or two of your attempts, to provide a starting point for discussion.

Spoiler: Hint

Consider integer powers of e.

?

 

[Dr.Peterson]

Hello!

I´m at my wits end with this problem. I´ve tried solving it like 10 times over multiple days. Can anyone nudge me in the right direction because I´m obviously missing some crucial knowledge to solve it.

$$\sqrt{\ln x}=\ln\sqrt{x}$$
Cheers!

I would start by simplifying so both sides are expressed in terms of $\ln(x)$; then it might help to do a substitution, letting $y = \ln(x)$. It will be easier to solve for $y$.

 

[Deleted member 4993]

Hello!

I´m at my wits end with this problem. I´ve tried solving it like 10 times over multiple days. Can anyone nudge me in the right direction because I´m obviously missing some crucial knowledge to solve it.

$$\sqrt{\ln x}=\ln\sqrt{x}$$
Cheers!

Consider:

ln(a^m) = m * ln(a)

 

[pka]

I´m at my wits end with this problem. I´ve tried solving it like 10 times over multiple days. Can anyone nudge me in the right direction because I´m obviously missing some crucial knowledge to solve it.: $$\sqrt{\ln x}=\ln\sqrt{x}$$

From the statement of the question it is necessary that $x>1$ otherwise $\sqrt{\log(x)}$ is not defined.

$\sqrt{\log(x)}=\frac{1}{2}\log(x)\\2\sqrt{\log(x)}=\log(x)\\4(\log(x))=(\log(x))^2$
Can you finish? If not please ask questions.

 

[lookagain]

From the statement of the question it is necessary that $x>1$ otherwise $\sqrt{\log(x)}$ is not defined.

$\displaystyle 4(\log(x))=(\log(x))^2$

It is necessary that $\displaystyle \ x \ge 1. \$ $\displaystyle \ \ \sqrt{\log(x)} \ = \ 0 \$ when x = 1.

$\displaystyle 4(\log(x))=(\log(x))^2 \$ will provide both solutions.