Question

Rearranging Help... Quite Hard!

Answer 1

Show:
\[\frac{ -1 }{ k }\ln(80-kV) = t-\frac{ 1 }{ k }\ln80\]

may be expressed as
\[V=\frac{ 1 }{ k }(80-80e^{-kt})\]

Answer 2

I have tried putting t as
ln e^t and using the log laws but I'm getting the wrong answer...

Answer 3

add (1/k ln 80) on each side

Answer 4

Yeah but how would we simplify that?
\[\frac{ -1 }{ k } \ln(80−kV)+\frac{ 1 }{ k } \ln(80)\]

Answer 5

@gorv

Answer 6

=t
on right side

Answer 7

now 1/k we can take common on left side

Answer 8

and using property of log
log a - logb=log(a/b)

Answer 9

Ok so\[\frac{ 1 }{ k }(\ln(80)-\ln(80-kV)) = \frac{ 1 }{ k }(\ln\frac{ 80 }{ 80-kV })\]

Answer 10

yep

Answer 11

which is equal to
ln e^t on the other sides I presume?

Answer 12

\[\frac{ 1 }{ k }\ln \frac{ 80 }{ 80-kv }=t\]

Answer 13

no we will multiply by k on both side frist

Answer 14

\[\log_{a} b=c\]
by using log property
\[b=a^c\]

Answer 15

a= base of log

Answer 16

and base on ln = e

Answer 17

Can you show me how to do that I'm not too sure
1/k (80/80-kv) = e^t ???

Answer 18

how u got e^t

Answer 19

\[\frac{ 1 }{ k }*\ln \frac{ 80 }{ 80-kv }=t\]

Answer 20

multiply k by both side

Answer 21

\[\ln \frac{ 80 }{ 80-kv }=kt\]

Answer 22

ln has base =e

Answer 23

\[\frac{ 80 }{ 80-kv }=e^{kt}\]

Answer 24

divide both side by e^kt

Answer 25

and multiply by 80-kv

Answer 26

\[\frac{ 80 }{ e^{kt} }=80-kv\]

Answer 27

\[80*e^{-kt}=80-kv\]

Answer 28

subtract 80 from both side

Answer 29

\[80e^{-kt}-80=-kv\]

Answer 30

divide both side by k

Answer 31

\[\frac{ 1 }{ k }(80e^{kt}-80)=-v\]

Answer 32

multiply by -1 on both side

Answer 33

\[\frac{ -1 }{ k }(80e^{-kt}-80)=v\]

Answer 34

\[\frac{ 1 }{ }(80-80e^{-kt})=v\]

Answer 35

I see now! Thanks gorv, so much rearranging...