Question

Can someone help me solve for x and check A= Be^rx ?

Answer 1

better equation \[A = Be ^{rx}\]

Answer 2

so we need to isolate x
start by dividing B on both sides

Answer 3

okay now i have \[A/B=e ^{rx}\]

Answer 4

yes,
you can now take natural logarithm (ln) on both sides.

Answer 5

on the right side, use the property that
\(\ln A^B = B \ln A\)

Answer 6

would it be \[\ln(A/B) =rx \ln e\] ??

Answer 7

@hartnn will i get
\[\ln(A/B)=rx \ln e\] ??

Answer 8

yes!
and ln e = 1

Answer 9

then you just divide both sides by 'r'

Answer 10

my final answer is \[x= \ln (A/B) /r\]

Answer 11

and it is correct :)

Answer 12

do you perhaps know how i would check my answer?

Answer 13

check means to confirm whether it is correct ?
its correct!

Answer 14

or u want to bring the original equation back from your answer ?

Answer 15

yeah but my teacher actually wants poof that the answer is correct

Answer 16

so lets bring original equation back from your answer
do the reverse steps
multiply 'r' on both sides

Answer 17

wouldn't i have to plug in by putting \[A=Be ^{r(\ln(A/B))/r}\]

Answer 18

ok, yes you can do that too
see whether you simplify right side = A

Answer 19

yeah i think so what i did was\[A=Be ^{\ln(A/B)}\] since the r's canceled
then i thought \[e ^{\ln}\] cancel too then i'm left with \[A=B ^{A/B}\] then i crossed the b's and i'm left with\[A=A\] ... does that seem right??

Answer 20

yes, but its not like e^ln "cancels" out
its the use of the property that
\(\large a^{\log_ax} = x \)

Answer 21

so, \(\large e^{\ln {(stuff)} }=stuff\)

Answer 22

everything elase was correct :)

Answer 23

so my check needs to have the log equation you wrote down?

Answer 24

when you say
e^(ln A/B) = A/B
just mention this in the reason besides it,
a^ log_a x =x

Answer 25

oh okay Thank you for your help, I really appreciate it :)

Answer 26

you're welcome ^_^