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# C3 logs help! watch

1. mighty stuck on this whole exponential logs stuff, any ideas how to solve this to get exact solutions?

lnx + ln3x = 27

any help muchly appreciated
2. can you remember the log rules, when you have logs of the same base and add them then you can multiply what is being logged (remember?)
does this help?
3. (Original post by thebassking)
can you remember the log rules, when you have logs of the same base and add them then you can multiply what is being logged (remember?)
does this help?
i think so, got as far as

ln3x^2 = 27, and then ground to a halt there :/

would you bring the 2 down to make 2ln3x = 27?
4. (Original post by periwinkle304)
i think so, got as far as

ln3x^2 = 27, and then ground to a halt there :/

would you bring the 2 down to make 2ln3x = 27?
yep, carry that on now to try and get a result for x, i.e. ln3x^2 = 27/2 ... then do the exponential both sides? Then you should get a value for x. what do you get?
5. (Original post by thebassking)
yep, carry that on now to try and get a result for x, i.e. ln3x^2 = 27/2 ... then do the exponential both sides? Then you should get a value for x. what do you get?
right, so i ended up with ln3x = 13.5

3x = e^13.5
so x = (e^13.5)/3

but when i subbed that back into the original equation, i got 25.9 instead of 27?

ugh its late, i give up lol, thanks so much for your help anyway
6. (Original post by periwinkle304)
right, so i ended up with ln3x = 13.5

3x = e^13.5
so x = (e^13.5)/3

but when i subbed that back into the original equation, i got 25.9 instead of 27?

ugh its late, i give up lol, thanks so much for your help anyway
you forgot that it is 3x^2 = e^13.5
That should set you right
No problem!

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