Question

the point P(a, klogea) lies on the curve y =klogex where "a" and "k" are constants
a) show that the equation of the tangent T to the curve at P is kx-ay +ak(logea-1)

Answer 1

is the base e?
or is that e*a?

Answer 2

\[P(a,k*\log_e(a)) or P(a,k*\log(e*a))\]?

Answer 3

\[y=k*\log_ex or y=\log(e*x)\] ?

Answer 4

lost

Answer 5

ok i will asumme it is base e and if it isn't correct oh well at least you will have an example

Answer 6

\[y=k\ln(x)\]?

Answer 7

\[y=k*lnx\]
\[y'=k*\frac{1}{x}\]
\[y'(a)=k*\frac{1}{a}\] is the slope at x=a

Answer 8

so point is
\[(a, k\ln(a))\] now that makes sense yes?

Answer 9

\[y=mx+b\]
\[k*lna=\frac{k}{a}*a+b\]
we need to solve for b the y-intercept
\[b=k*lna-k\]
so the tangent line at P is
\[y=\frac{k}{a}*x+k*lna-k\]

Answer 10

point slope formula give
\[y-k\ln(a)=\frac{k}{a}(x-a)\]

Answer 11

\[y=\frac{k}{a}x-k+k\ln(a)\]

Answer 12

does multiplying by a get what you want?

Answer 13

looks like we were suppose to get slope is k

Answer 14

so maybe we didn't interpret what she had correctly
oh well

Answer 15

i give you a medal satellite :)

Answer 16

want this
\[kx-ay +ak(\ln(a)-1)\]

Answer 17

she/he whatever

Answer 18

but where is the = sign

Answer 19

oh not it is ok. just algebra

Answer 20

wait is it y=kx-ay+ak(lna-1)

Answer 21

ikr maybe it is this
\[kx-ay =ak(\ln(a)-1)\]

Answer 22

i bet that is it.

Answer 23

multiply by a and we get it

Answer 24

the plus sign = equal sign
that would work

Answer 25

yes i think so. just
\[y=\frac{k}{a}x-k+k\ln(a)\]
\[ay=kx-al+ak\ln(a)\] and some other stupid algebra probably does it.

Answer 26

and don't make any... typos!

Answer 27

@myininaya and @satellite73 THANK YOU A BILLION TIMES OVER and i deeply apologise for the confusion. like myininaya had suggested, it is base e! THANK YOU, I WOULD GIVE YOU GUYS 10000000000000000000 MEDALS EACH IF THAT WAS POSSIBLE!

Answer 28

lol np