Question

If $1000 is invested at 5% compound monthly for 4 years the compound amount is given.by A=1000(1.005)^48. Given that log find log A

Answer 1

log both sides ....

Answer 2

How

Answer 3

?

Answer 4

A=1000(1.005)^48

log(A) = log (1000(1.005)^48)

and work whatever log properties you can remember

Answer 5

1000 = 10*10*10

log10 = 1

not too sure if the 1.005 is correct, but ill assume a certain degree of intelligence on it :)

Answer 6

It is

Answer 7

I still don't understand how to.get the answer

Answer 8

do you know your log properties?

Answer 9

If there are 10 months to a year, then 0.005 is correct.

As it is the case that there are 12 months in a year, you may wish to go with 0.05 / 12 = 0.004166667

Answer 10

the basics are:

log(ab) = loga + logb

log(a/b) = loga - logb

log(a^c) = c loga

Answer 11

Not really

Answer 12

and if we assume log is base 10 ... log(10) = 1

Answer 13

Yea

Answer 14

then its just a matter of applying those properties

log(A) = log (1000(1.005)^48)
= log (1000) + log(1.005)^48
= log (10^3) + log(1.005)^48
= 3 log (10) + 48 log(1.005)
= 3 + 48 log(1.005)

Answer 15

there is no simple way i can see to hand up that log 1.005

Answer 16

Now.how.do.we getvA?

Answer 17

you would under the log by base 10 ing both sides ....

Answer 18

Example?

Answer 19

\[log(A)=a+log(b)\]
\[\Large 10^{log(A)}=10^{a+log(b)}\]
\[\Large A=10^{a+log(b)}\]
\[\Large A=10^{a}~10^{log(b)}\]

Answer 20

i see one more move :)
\[\Large A=10^{a}~10^{log(b)}\]
\[\Large A=10^{a}~b\]

Answer 21

of course this just gets back to:

A = 1000 (1.005)^48

Answer 22

So what would the answer be

Answer 23

im not sure that youve made your objective clear enough to determine that ....

are you simply trying to determine the value of A? if so, what was the need to go thru all that log work in the first place?