Question

ln[\z+2\]-ln[\z+3\]=ln[\t\]+c
then, i apply e^ on both side
e^(ln[\(z+2)/(z+3)\]=e^(ln[\t\]+c)
=[\(z+2)/(z+3)\] = [\(t)(e^c)\] where e^c =k
[\(z+2)/(z+3)\]=[\(k t)\]

is this right???

or

e^(ln[(z+2)/(z+3)] =e^(ln t)+e^c
then
(z+2)/(z+3)=t+K where K=e^c

which one is right??

Answer 1

\[ at the beginning

Answer 2

you mean first one is right?

Answer 3

for latex ... use \[ at the beginning

Answer 4

oops. thank you !

Answer 5

Yes, the first method you applied was correct. :)

Answer 6

thank you! can u explain why is like that?? i mean why i cant e to the each lnt and c

Answer 7

You have to exponentiate the `entire side`.

Which gives,\[\Large e^{\left(\ln t+c\right)}\]A handy rule of exponents tells us:\[\Large a^{b+c} \quad=\quad a^b\cdot a^c\]

Answer 8

\[\Large e^{\left(\ln t+c\right)} \quad=\quad e^{\ln t}\cdot e^{c}\]

Answer 9

Why can't you exponentiate them individually?
I'm not sure why.. I don't know if there exists an operation that allows us to do that :o hmm

Answer 10

thank you!!! i was just arguing about this with gf , i was keep telling her have to exp whole part. but just cant explain why..

Answer 11

The same applies in reverse.
If you had something like,\[\Large stuff\quad=\quad e^t+C\]You wouldn't be able to apply the log to each term individually.\[\Large \ln(stuff) \quad=\quad \ln(e^t+C)\]

Answer 12

In ^ that particular example you would probably want to move the C to the other side so you could do sneaky stuff to the e^t :)\[\Large stuff-C=e^t\]But whatev, it's just an example ^^ heh

Answer 13

thank you very much, very clear explanation.