help me with this problem 2-ln(3-x)=0 can u show step by step?

Answer 1

2-ln(3-x)=0
ln(3-x)=2
e^ln(3-x)=e^2
3-x=e^2
x=3-e^2

Answer 2

step one: add ln(3-x) to both sides
2 = ln(3-x)
step two: exponentiate e to both sides (in other words, take base e anti-log of both sides)
e^2 = e^ln(3-x) = 3-x
step three: add (x-e^2) to both sides
x = 3-e^2
optional step four: use a calculator to get an approximation
x=-4.389


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answer is negative
below, "tinned" made a mistake in the step that gave him
3-e^2 = x (it should be 3-e^2 = -x)
guess that "AS level maths" isn't working out so well

Answer 3

ln (3 - x) = 2
3 - x = e ²
x = 3 - e ²

Answer 4

step one: add ln(3-x) to both sides
2 = ln(3-x)
step two: exponentiate e to both sides (in other words, take base e anti-log of both sides)
e^2 = e^ln(3-x) = 3-x
step three: add (x-e^2) to both sides
x = 3-e^2
optional step four: use a calculator to get an approximation
x=-4.3890

Rounds to 4

Hope I helped!

Easy... Hey bro... Ur in Grade 5 right?

Answer 5

2 - ln(3-x) = 0

so,

2 = ln(3 -x)

e^2 = 3-x

(e ^2) -3 = x

4.389056... = x

x = 4.39 (3 significant figures)

Answer 6

2-ln(3-x)=0

2=ln(3-x)

e^2 = 3-x

3-2.718*2.718=x (approx)

or -4.38905609893068

Answer 7

Ahh... No!

Answer 8

Its cows!!!!

Answer 9

no i cant

Answer 10

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