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Mathematics 19 Online
OpenStudy (baldymcgee6):

How do I simplify ln(ln(2)+ sqrt((ln(2))^2+1))

OpenStudy (baldymcgee6):

\[\ln(\ln(2)+ \sqrt((\ln(2))^2+1))\]

OpenStudy (baldymcgee6):

@Algebraic! i'm sure you have a solution.. :)

OpenStudy (anonymous):

sure it's written correctly?

OpenStudy (baldymcgee6):

i'm evaluating inverse sinh at ln(2)

OpenStudy (anonymous):

maybe its like this? ln[(ln(2)+ sqrt((ln(2))^2+1))]

OpenStudy (baldymcgee6):

\[\ln(\ln(2)+ \sqrt((\ln(2)^2+1))\]

OpenStudy (anonymous):

ln[(ln(2)+ sqrt((ln(2))^2+1))]= 0.64704348 inverse of sinh (ln2)= 0.64704348

OpenStudy (baldymcgee6):

supposed to have an exact value.

OpenStudy (anonymous):

doesn't

OpenStudy (anonymous):

is it ln (2^2) ???

OpenStudy (anonymous):

or (ln(2))^2

OpenStudy (baldymcgee6):

(ln(2))^2, my bad.

OpenStudy (baldymcgee6):

\[\ln(\ln(2)+ \sqrt((\ln(2))^2+1))\]

OpenStudy (baldymcgee6):

I don't think it can be simplified anymore further. I thought you could square root the (ln(2))^2 + 1 to give just ln2 +1

OpenStudy (anonymous):

so the root is on... (ln2)^2 +1 ..

OpenStudy (baldymcgee6):

\[\sqrt{(\ln2)^2 +1} = 2\ln2 +1\]

OpenStudy (baldymcgee6):

??

OpenStudy (anonymous):

no...

OpenStudy (baldymcgee6):

@Algebraic! \[\sqrt{(\ln2)^2 +1} = ??\]

OpenStudy (baldymcgee6):

@satellite73 any help?

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