Time payments (1 Viewer)

[Linda N]

I got trouble simplifying this working out below to find n.

1,200,000 (1.005)^n = 8000 [1 (1.005)^n - 1 / 1.005 - 1]

(1.005)^n = 4 ----> this section i got trouble.

 

[cAinesy]

what is the question.. and ill show u how to do it

 

[Brad]

Use logs....

 

[Brad]

as in (1.005)^n = 4
Log(1.005)^n= log 4
nlog1.005=log4
n=log4/log1.005

 

[smallcattle]

(1.005)^n = 4

n= ln 4 / ln 1.005

 

[Linda N]

Sorry.. i meant

I don't know how to get from this:
1,200,000 (1.005)^n = 8000 [1 (1.005)^n - 1 / 1.005 - 1]

to (1.005)^n = 4


The question is:
Maree is retiring nextweek and her Superannuation Fund contains $1,200,000. The fund is earning 6% p.a compound interest, compounding monthly. Maree wishes to withdraw a regular of $8000 per month to live on in her retirement.

By finding a similar expression for the amound remaining after n months, find how many years the money will last.

 

[smallcattle]

1,200,000 (1.005)^n = 8000 [1 (1.005)^n - 1 / 1.005 - 1]

150 (1.005)^n = 1 (1.005)^n - 1 / 0.005

3/4 (1.005)^n = (1.005)^n - 1

3/4 (1.005)^n - (1.005)^n = -1

-1/4 (1.005)^n = -1

(1.005)^n = 4

 

[Linda N]

Hehe thanx. I realise i done a silly mistake.

 

[Sarah168]

Hey Linda

I did it, but in another way so I dont end up with -1, but the final answer was like 277 years! wtf... I always get thrown off by "wrong -sounding" answers...ln4/ln1.005 is 277 right?