Question

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

Answer 1

5ln(x-2) = 30

Answer 2

ln(x-2) = 6

Answer 3

ln or log ?

Answer 4

\( 5ln(x-2) = 30 \)

\(ln(x-2) = 6\)

\(log_e(x-2) = 6\)

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

Answer 5

ln

Answer 6

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

Answer 7

bring the \(2\) to the RHS

Answer 8

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

Answer 9

+2

Answer 10

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

Answer 11

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

Answer 12

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

Answer 13

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

Answer 14

yea

Answer 15

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

Answer 16

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

Answer 17

so what do you do next

Answer 18

well i just told you before :)

Answer 19

bring the 2 to the RHS?

Answer 20

ooooohhhhhhhh
xD
duh x = e^6 + 2

Answer 21

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

Answer 22

i forgot i was solving for x , silly me

Answer 23

i more last question what do you do for

e^x-3 = 16

Answer 24

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

Answer 25

e^x-3 + 4 =20

Answer 26

Um, that is not right..

Answer 27

drag the \(-3) to the rhs

Answer 28

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

Answer 29

\[ e^x = 16+ 3\]

Answer 30

\[ e^x = 19\]

Answer 31

Take ln both sides
\[xlne = ln19\]

Answer 32

the answer is ln(16)+3

Answer 33

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

Answer 34

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

Answer 35

my teacher was wron. she gave me that answer

Answer 36

well there is a low chance that wolfram alpha is wrong

Answer 37

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?
\[e ^{x-3} = 16\]