Question

y=In3x, where x=e
(equation of line)

Don't know how to work with e.
Differentiations = 1/x

and plugging in value gives 1/e
...How to carry on?

Answer 1

Wait, what are you trying to find/do?

Answer 2

Find the equation of the line.

Answer 3

well the derivative is just the slope at any point. The point is (e), remember (e) is a constant, so 1/e is a real number.

Answer 4

knowing that, you just need to find the rest of the tangent line:

f(x) - f(e) = (1/e)(x - e)

Answer 5

so just plug (e) back into the original function:

ln(3e) = ln3 + ln(e) = ln3 + 1

Answer 6

the whole thing (i'm assuming) i think should look like:

y - (ln3 + 1) = (1/e) (x-e)

someone correct me if i'm off.

Answer 7

The answer to this whole thing is ey=x+eIn3
I just don't know how to get it.

Answer 8

1st find y when x = e ln(3e) = ln(3) + 1 since ln(e) = 1

then dy/dx = 3/3x or dy/dx = 1/x to findthe gradient substitute x = e

then m = 1/e and point (e, (ln(3)+1))

\[y - (\ln(3)+1) = 1/e(x - e)\]

y = x/e - ln(3)

Answer 9

going off my equation:

y - ln3 - 1 = (1/e) (x-e)

multiply both sides by (e)

ey - eln3 -e = x -e

ey = x - e + e + eln3

ey = x + ln3

Answer 10

hint: y prime=slope.

given the x or the x intercept we can find the equation of the line y at 0.

Answer 11

for your answer order... they multiplied through by e to get rid of the fraction

Answer 12

ey = x + eln3* sorry made a typo :)

Answer 13

Oh. Ok. I need to look at this again. Thanks guys!

Answer 14

small problem jerwyn.... you need to substitute x = e to find the value of y ... don't assume