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solve without using calculator 5ln(x-2)-2=28 help please, thank you

Mathematics
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5ln(x-2) = 30
ln(x-2) = 6
ln or log ?

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Other answers:

\( 5ln(x-2) = 30 \) \(ln(x-2) = 6\) \(log_e(x-2) = 6\) \((x-2) = e^{6}\) Are you able to do it now?
ln
im stuck at the (x-2)=e^6
bring the \(2\) to the RHS
i know the answer is e^6 +2 but i dont know where the =2 comes from
+2
Because when you bring the \(-2\) to the RHS it's a positive
I don't know the correct way to say that lol
alright can you walk me through the part after ln(x-2)=6
Well you know that \(ln\) can also be written as \(log_e\) right?
yea
so you get e^6= (x-2) right
yeah \((x-2) = e^{6}\)
so what do you do next
well i just told you before :)
bring the 2 to the RHS?
ooooohhhhhhhh xD duh x = e^6 + 2
yes; you dont really bring it was called something else which forgot. just simply drag it to RHS. But there is a better name for it lol
i forgot i was solving for x , silly me
i more last question what do you do for e^x-3 = 16
\[ e^x - 3 =16\] What have you tried?
e^x-3 + 4 =20
Um, that is not right..
drag the \(-3) to the rhs
but its an exponent x-3 is the exp. for e
\[ e^x = 16+ 3\]
\[ e^x = 19\]
Take ln both sides \[xlne = ln19\]
the answer is ln(16)+3
Wolfram Alpha agrees with me http://www.wolframalpha.com/input/?i=+e%5Ex-3+%3D+16
my teacger myst be wring though . she gave me that answer
my teacher was wron. she gave me that answer
well there is a low chance that wolfram alpha is wrong
oh, well, it looks like we have a paranthesis confusion here. when writing e^x-3 = 16, it's important that you write the parantheses aroudnthe (x-3), else we can't know that the whole thing is the exponant. So we have this right? \[e ^{x-3} = 16\]

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