Question

Topic: Base e and Natural Logarithms
ln (x-2) = 2
and another one is
ln x + ln 2x = 2

Answer 1

Hint*

\[\huge e^{ln(x - 2) = (x - 2)}\]

Answer 2

\[x-2=e^2,x=?\]

Answer 3

I'm confused :(

Answer 4

\[\ln a+\ln b=\ln ab\]

Answer 5

so ln 2x^2

Answer 6

Well lets tackle question 1...you said you were confused....what on?

Answer 7

all of it because I thought e was ln

Answer 8

e is the inverse of ln

Hence...when you raise e^ln power....they cancel out...that is key in logarithm problems

\[\large ln(x - 2) = 2\]

raise e to both sides

\[\huge e^{ln(x - 2)} = e^2\]

\(\large e^{ln} = 1\) so we have

\[\large x - 2 = e^2\]

now you would just solve for 'x'

Answer 9

so I would basically just solve x-2=2 fir x

Answer 10

for

Answer 11

Well

\[\large x - 2 = e^2\] for 'x' yes (you forgot the 'e' above\]

Answer 12

so add 2 right

Answer 13

Correct

Answer 14

but then how would I solve x=2e^2

Answer 15

put it in my calculator

Answer 16

you mean x = e^2 + 2 right?

Answer 17

ohhhh ok sry my bad

Answer 18

so 2.6931 = x

Answer 19

Not quite what I get....was that out of your calculator?

Well put it like this...

e is approximately equal to 2.718 (If I remember correctly

so we have

\[\large x = 2.718^{2} + 2\]

what does that come out to?

Answer 20

9.3875 and I would plug 2.718 to any problem similar to this one

Answer 21

Well that is just approximate...if you want the exact answer..use the 'e' key in your calculator...

But yes you can do that

Answer 22

ok and umm do you mind helping me with one more problem.