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A solution to a problem here is unclear to me :

The problem:

How long does it take $\$100$ to become $\$1000$ if invested at $10\%$ compounded quarterly?

His solution:

$A_0 = 100, A(t) = 1000, r = .1, n = 4$ \begin{align*} 1000 & = 100\left(1 + \frac{.1}{4}\right)^{4t}\\ 10 & = 1.025^{4t} && \text{use the change of base formula}\\ \log 10 & = \log 1.025^{4t}\\ 1 & = 4t\log 1.025 && \text{$\log 10 = 1$}\\ \frac{1}{4\log 1.025} & = t\\ t & = 23.31 \end{align*} It will take $23.3$ years to have $\$1000$ from the $\$100$ investment.

He says "Use the change of base formula". But where does the "change of base formula" happen?

I know the change of base formula is $$\log_a b = \frac{\log b}{\log a},$$ but I don't see it in his solution.

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    $\begingroup$ Possible typos: line 2 should be "1000 = 100(1 + .1/4)^(4t)", line 3 should be "10 = 1.025^(4t)", line 4 should be "Log 10 = Log 1.025^(4t)" ? $\endgroup$ – Patrick Hew Sep 6 '18 at 5:49
  • $\begingroup$ Welcome to MathSE. Please read this MathJax tutorial, which explains how to typeset mathematics on this site. $\endgroup$ – N. F. Taussig Sep 6 '18 at 7:19
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The change of base formula is implicitly applied between lines 4 and 5. The question is "What is $\log 1.025^{4t}$?". The change of base formula allows us to do the following: $$\begin{align*} \log_{1.025} 1.025^{4t} &= \frac{ \log 1.025^{4t} }{ \log {1.025}} \\ 4t &= \frac{ \log 1.025^{4t} }{ \log {1.025}} \end{align*}$$ hence $$ \log 1.025^{4t} = (\log {1.025}) 4t $$

I suspect that you didn't see it because you instinctively knew that $\log 1.025^{4t} = 4t \log 1.025$ under the shortcut $\log a^b = b\log a$, but the logic behind that shortcut is the change of base formula.

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  • $\begingroup$ But how do you guess that it's using base 1.025 in the first place ? $\endgroup$ – trogne Sep 8 '18 at 17:01
  • $\begingroup$ I see. You get the known "log 1.025^4t" , and use it inside a change of base formula... $\endgroup$ – trogne Sep 8 '18 at 18:05

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