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Given:

$f(t)=500(1.1)^t$

How would I mathmatically solve for:

$f^{-1}(3000)$

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    $\begingroup$ Solve $3000=500(1.1)^t$. First divide both sides by $500$, then take logs... $\endgroup$ Commented Sep 8, 2013 at 23:16
  • $\begingroup$ @DavidMitra That would be for $f(t)$, not $f^{-1}(t)$. $\endgroup$
    – Don Larynx
    Commented Sep 8, 2013 at 23:18
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    $\begingroup$ No @DavidMitra is right . $\endgroup$
    – what'sup
    Commented Sep 8, 2013 at 23:19
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    $\begingroup$ @Jossie $f^{-1}(3000)$ is the value $t$ for which $f(t)=3000$. $\endgroup$ Commented Sep 8, 2013 at 23:19
  • $\begingroup$ @Jossie $ f(t) = 500(1.1)^t \Rightarrow f^{-1} ( 500(1.1)^t ) = t $ $\endgroup$
    – what'sup
    Commented Sep 8, 2013 at 23:20

2 Answers 2

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An easier suggestion: \begin{align*} &3000=500(1.1)^{t} \Longrightarrow\\ &6 = (1.1)^{t} \Longrightarrow\\ &\ln{6} = t \ln{1.1} \Longrightarrow\\ &t = \frac{\ln{6}}{\ln{1.1}}\Longrightarrow\\ \end{align*} So if we have the coordinate pair $\left(\frac{\ln{6}}{\ln{1.1}}, 3000 \right) \in f$, then
$\left(3000,\frac{\ln{6}}{\ln{1.1}}\right)\in f^{-1}$
Yielding $$f^{-1}(3000) = \frac{\ln{6}}{\ln{1.1}}$$

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1) Reverse $f(t)$ and $t$

$t = 500(1.1)^{f(t)}$

2) Solve for $f(t)$ (which is really now $f^{-1}(t)$)

Hint: use logarithms

3) Plug in $f^{-1}(3000)$.

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