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The compound interest formula is

$$ A =P( 1+r ) ^{n} $$

If I want to find $n$ it seems logical to that I just solve for it using the logarithm base (1+r), like so:

First divide by $P$:

$$ \dfrac{A}{P} = ( 1+r) ^{n} $$

and then:

$$ log_{(1+r)}(\dfrac{A}{P}) = n $$

which is easily solvable. However, this is wrong, apparently. It seems logical to me that you can just apply the rule that if $b^x=y$ then $log_{b}(y)=x$. Can somebody explain why this does not work? Thanks a lot!

EDIT: I made a mistake - I meant "solve for $n$", not "solve for $r$", as I wrote first. Sorry for the confusion!

jk47
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    It's not strictly wrong, but it doesn't help you solve for $r$. The only way to get $r$ from here is to ... raise both sides to the power of $r$ again. Instead of taking a logarithm, take the $n$th root instead. – PrincessEev Dec 07 '23 at 13:08
  • ‘However, this is wrong, apparently’ - Why? – Soham Saha Dec 07 '23 at 14:19
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    What you did is technically correct. However, what we would normally do is take $\ln$ on both sides, so you get $$ \ln\left(\frac{A}{P}\right) = n\ln(1 + r) \implies n =\frac{ \ln\left(\frac{A}{P}\right)}{\ln(1 + r)} $$

    I don't know what you mean by "wrong" though.

     – Pavan C. Dec 07 '23 at 14:57
  • You aren't wrong. The change-of-base formula $\log_b(x)=\frac{\ln(x)}{\ln(b)}$ would turn your answer into the same answer in the previous comments $n=\frac{\ln(\frac{A}{P})}{\ln(1+r)}$. Many people would frown on a log with base 1+r because most calculators can only easily handle base 10 (log) or base e (ln). – J. Chapman Dec 07 '23 at 17:05

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