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Mathematics 14 Online
OpenStudy (anonymous):

log(ln(ex))=2... is the answer x=e^99??

OpenStudy (anonymous):

is it Log(Ln(e^x)) =2 so no.. this is not the answer

OpenStudy (anonymous):

how do you simplify Ln(e^x) ?

jhonyy9 (jhonyy9):

so do you know what is the base of ln ?

jhonyy9 (jhonyy9):

ln is logarithm natural ,yes ?

jhonyy9 (jhonyy9):

corage ,what is the base of ln ?

jhonyy9 (jhonyy9):

courage wann being there ,sorry

jhonyy9 (jhonyy9):

so if the base of logarithm naturale is e so than lne = ? can you answer this ?

OpenStudy (anonymous):

it's not log(ln(e^x))=2 @Coolsector :D lne= 1 right?

OpenStudy (anonymous):

yes lne = 1

OpenStudy (anonymous):

and ln(e^x) = .. ?

OpenStudy (anonymous):

and can you please make the question more clear ? i dont really understand what is the question if it is not what i wrote

OpenStudy (anonymous):

it's not e^x :D it's really like that :) log(ln(ex))=2 if you want \[\log(\log_{e} (ex))=2\]

OpenStudy (anonymous):

ok so log(ln(e) + ln(x)) = 2 log(1+ln(x)) = 2 1 + ln(x) = 100 ln(x) = 99 x = e^99 and you were right at the first place

OpenStudy (anonymous):

so sorry for that :)

OpenStudy (anonymous):

yay thanks! i just needed to double check:D

OpenStudy (anonymous):

log(loge(ex))=2 (loge(ex) = 100 ex = e ^ 100 x = e ^ (100- 1) x = e ^99

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