Question

4 = 8(1 - 1.03^n)/(-0.03)

Solve for n

Answer 1

\[4 = 8(1 - 1.03^{n}) / (-0.03)\]

Answer 2

What would you get if you multiply by -0.03 to both sides?

Answer 3

\[\large 4*(-0.03) = 8(1-1.03^n)\]\[\large -0.12 = 8(1-1.03^n)\]
Then you divide by 8 to both side
\[\large \frac{-0.12}{8} = \frac{8(1-1.03^n)}{8}\]
That would give
\[\large \frac{-3}{200} = 1-1.03^n\]
Then send 1 to the left
\[\large 1- \frac{-3}{200} = 1.03^n\]
That would be \[\large \frac{203}{200} = 1.03^n\]

What I did above it basically isolate \[\large 1.03^n\] so it would become an easy logarithm question

Next step would be
\[\huge \log_{1.03}\frac{203}{200} = n\]

Answer 4

I have no clue lol

Answer 5

I don't even know logarithms, they want to to use "Trial and Error" to find the value of n.

My teacher told me it is a nice two digit number.

Answer 6

Two digits? more like 0.5036947158...etc lol

Answer 7

looks like you did the steps correctly zepp

Answer 8

Then what about syaing that the answer is \[0<n<1\]?

Answer 9

but you could simplify the last step as follows:\[\frac{203}{200}=1.03^n\]take logs of both sides:\[\log(\frac{203}{200})=\log(1.03^n)=n\log(1.03)\]therefore:\[n=\frac{\log(\frac{203}{200})}{\log(1.03)}\]

Answer 10

to avoid log to base 1.03

Answer 11

answer comes out at roughly 0.5

Answer 12

hang on - just read the response from the asker - they want to use trial and error not logs

Answer 13

@Leaper if you are familiar with the Newton-Raphson Method, then you could use:\[4=8(1 - 1.03^n)/(-0.03)\]and re-arrange to get:\[f(n)=4-8(1 - 1.03^n)/(-0.03)\] such that we want to find a value for n that makes f(n) zero. the N-R Method states:\[n_{k+1}=n_k-\frac{f(n_k)}{f'(n_k)}\]let me know if you are aware of this method and I will explain this further.