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Inx^2-In36=Ine
Original post by Cameron45
Inx^2-In36=Ine


What about it?
Reply 2
Trying to solve for x but don't know where to start..
Original post by RDKGames
What about it?
Original post by Cameron45
Trying to solve for x but don't know where to start..


Plenty of ways to go about it.

You can add ln36\ln 36 to both sides then combine RHS into a single log by using the rule loga+logb=log(ab)\log a + \log b = \log(ab)
remember that ln(e) has a special value :h:
Reply 5
The answer is 6. I've tried adding In36 to the other side but I'm not getting an answer of 6
Original post by RDKGames
Plenty of ways to go about it.



You can add ln36\ln 36 to both sides then combine RHS into a single log by using the rule loga+logb=log(ab)\log a + \log b = \log(ab)
Original post by Cameron45
The answer is 6. I've tried adding In36 to the other side but I'm not getting an answer of 6


That's not a correct answer.

ln(62)ln(36)=ln(36)ln(36)=0lne\ln(6^2) - \ln(36) = \ln(36)-\ln(36) = 0 \neq \ln e
Reply 7
Oh. It says here that the answer is 6.


Original post by RDKGames
That's not a correct answer.


ln(62)ln(36)=ln(36)ln(36)=0lne\ln(6^2) - \ln(36) = \ln(36)-\ln(36) = 0 \neq \ln e
Original post by Cameron45
Inx^2-In36=Ine


ln x^2 - ln 36 = ln e
ln e = 1
2ln x= 1 ln 36
x = e^((1 ln 36)/2)
(edited 6 years ago)
Original post by brainmaster

ln e = 1
ln x^2 = ln 36


Note quite. You're basically claiming that 1+ln36=ln361+\ln 36 = \ln 36

Anyway, no full solutions. :smile:
Original post by RDKGames
Note quite. You're basically claiming that 1+ln36=ln361+\ln 36 = \ln 36

Anyway, no full solutions. :smile:


I deities my post thanks for correcting that I was multiplying instead of adding and I forgot about full solution I'll take care next time :smile:

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