Question

Find the exponential function f(x) = Ca^x whose graph goes through points (0,3) and (2,12).

Answer 1

hey there

Answer 2

HI!!

Answer 3

if the graph goes through that point then we can say that f(0)=3

Answer 4

oh @xapproachesinfinity got this right?

Answer 5

i guess @misty1212 lol

Answer 6

so urena do we have an agreement about what i just wrote ?

Answer 7

yeah, just say alex not my last name xp

Answer 8

ok i assumed it wasn't your real name sorry :)
so we can say f(0)=3
and f(2)=12
yes

Answer 9

aha its cool, and yass

Answer 10

ok good
now we see f(0) another by using that weird exponential function they give us
\[f(0)=Ca^0=C\]
similarly
\[f(2)=Ca^{12}\]
yes?

Answer 11

yeah i see how you got that

Answer 12

alright!
now since we have f(0)=3 and f(0)=C then C=3 yes?

Answer 13

yass

Answer 14

now we found C
we know that f(2)=12, and f(2)=Ca^12=3a^2
then 3a^2=12 yes

Answer 15

i thought it would be 3a^12 = 2

Answer 16

oh i did and error before after i said similarly
\[f(2)=Ca^2 \] a^2 not a^12 did a typo there sorry

Answer 17

ok i get it

Answer 18

yes my mistake, i was evaluating 2 but put 12 unwillingly

Answer 19

ok now \[3a^2=12\] is equation we can solve for a

Answer 20

any ideas how to solve it?

Answer 21

divide by 3

Answer 22

ok good start
a^2=4
what is next

Answer 23

square both sides so that a cancels and you have 4 and -4

Answer 24

hmm you mean square root both sides

Answer 25

yeah my bad xp

Answer 26

\[a^2=4 \Longrightarrow a=\pm \sqrt{4}\]

Answer 27

2 and -2

Answer 28

so a=2 or a=-2
yes

Answer 29

good! now we have to decide which is valid

Answer 30

how?

Answer 31

what did you learn about exponential function
don't they tell all the time a>1 yes?

Answer 32

right, so a=2

Answer 33

thank you so much

Answer 34

well a can be less than 1 too but not negative
since that can cause problems

Answer 35

ok note-to-self

Answer 36

so what they actually say is \[0<a<1~~or ~~a>1\]

Answer 37

you are welcome~

Answer 38

thanks for the testimony :)