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5ln(x-2) = 30

ln(x-2) = 6

ln or log ?

\( 5ln(x-2) = 30 \)
\(ln(x-2) = 6\)
\(log_e(x-2) = 6\)
\((x-2) = e^{6}\)
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

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