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

- anonymous

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

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- anonymous

5ln(x-2) = 30

- anonymous

ln(x-2) = 6

- anonymous

ln or log ?

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## More answers

- Mimi_x3

\( 5ln(x-2) = 30 \)
\(ln(x-2) = 6\)
\(log_e(x-2) = 6\)
\((x-2) = e^{6}\)
Are you able to do it now?

- anonymous

ln

- anonymous

im stuck at the (x-2)=e^6

- Mimi_x3

bring the \(2\) to the RHS

- anonymous

i know the answer is e^6 +2 but i dont know where the =2 comes from

- anonymous

+2

- Mimi_x3

Because when you bring the \(-2\) to the RHS it's a positive

- Mimi_x3

I don't know the correct way to say that lol

- anonymous

alright can you walk me through the part after ln(x-2)=6

- Mimi_x3

Well you know that \(ln\) can also be written as \(log_e\) right?

- anonymous

yea

- anonymous

so you get e^6= (x-2) right

- Mimi_x3

yeah \((x-2) = e^{6}\)

- anonymous

so what do you do next

- Mimi_x3

well i just told you before :)

- anonymous

bring the 2 to the RHS?

- anonymous

ooooohhhhhhhh
xD
duh x = e^6 + 2

- Mimi_x3

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

- anonymous

i forgot i was solving for x , silly me

- anonymous

i more last question what do you do for
e^x-3 = 16

- Mimi_x3

\[ e^x - 3 =16\]
What have you tried?

- anonymous

e^x-3 + 4 =20

- Mimi_x3

Um, that is not right..

- Mimi_x3

drag the \(-3) to the rhs

- anonymous

but its an exponent
x-3 is the exp. for e

- Mimi_x3

\[ e^x = 16+ 3\]

- Mimi_x3

\[ e^x = 19\]

- Mimi_x3

Take ln both sides
\[xlne = ln19\]

- anonymous

the answer is ln(16)+3

- Mimi_x3

Wolfram Alpha agrees with me
http://www.wolframalpha.com/input/?i=+e%5Ex-3+%3D+16

- anonymous

my teacger myst be wring though . she gave me that answer

- anonymous

my teacher was wron. she gave me that answer

- Mimi_x3

well there is a low chance that wolfram alpha is wrong

- anonymous

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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