Log Question (maths)

[sysD]

00 Ok, so here's the provided info:

 

A person invests $10 000 @ 3% per annum

Another person invests the same amount THREE YEARS LATER @ 5% per annum.

When will they be equal?

 

 

Here's the equation I came up with:

 

 

10 000 (1.03)^(n+3) = 10 000 (1.05)^(n)

 

This yields an answer of ~4.61

Here is my work:

 

10 000 (1.03)^(n+3) = 10 000 (1.05)^(n)

(n+3)log(1.03) = n(log(1.05))

n(log(1.03)) + 3(log(1.03)) = n(log(1.05))

3(log(1.03)) = n(log(1.05)) - n(log(1.03))

3(log(1.03)) = n(log1.05 - log1.03)

n = ( 3log1.03 ) / ( log1.05 - log1.03)

n = ~4.61

 

 

Here's the issue:

 

The book uses the equation:

 

10 000 (1.03)^(n) = 10 000 (1.03)^(n-3)

And here is the correct work and answer:

n(log(1.03)) = (n-3)(log(1.05))

n(log1.03) = n(log1.05) - 3log(1.05)

n(log1.03 - log1.05) = -3log(1.05)

n = (-3log(1.05)) / (log1.03 - log1.05)

n = ~7.61

 

This makes sense, as there are three less compounding periods (n-3) in the second (1.03%/annum) amount.

Why, though, is this not the same exact thing as saying the first amount has (n+3) three more compounding periods?

 

I understand that the maths yield a different answer, but would someone kindly conceptualy explain why the two are not equivilant?

[Schrödinger's hat]

There's nothing wrong with your logic, it's just what n represents.

Finish this sentence.

The balances will be equal 4.61 years after ________

Then this one

The balances will be equal 7.61 years after ________

 

This is why defining our terms explicitly is important, otherwise we can be talking about two different things without noticing.

[sysD]

Ahh, yes, I see.

 

Defining

"n" in this case as

"the time, in years, it would take for the second amount to equal the first"

would have circumvented this issue.

 

Alternatively, if I were to use the book's method,

"n" would be defined as:

"the time, in years, it would take for the first amount to equal the second."