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.

[math]\sqrt{\ln x}=\ln\sqrt{x}[/math]
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.

[math]\sqrt{\ln x}=\ln\sqrt{x}[/math]
Cheers!

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

 

[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.

[math]\sqrt{\ln x}=\ln\sqrt{x}[/math]
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.: [math]\sqrt{\ln x}=\ln\sqrt{x}[/math]

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

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

 

[lookagain]

From the statement of the question it is necessary that [imath] x>1[/imath] otherwise [imath]\sqrt{\log(x)} [/imath] 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.