3
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write the inverse of: $y= 3(4)^{2x+1} + 1$

This is what I did:

$ \begin{align} x=& 3(4)^{2y+1} + 1\\ x-1=& 3(4)^{2y+1} \\ \frac {(x-1)} 3=& (4)^{2y+1}\\ \log((x-1)/3)=& (2y+1)\log(4)\\ \log((x-1)/3)=& 2y\log(4)+\log(4)\\ y=&[\log((x-1)/3)-\log(4)]/2\log(4) \end{align} $

Did I do this correctly?

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  • $\begingroup$ At some point the $y$ became $x$ and then back again to $y$, but yes, the rest is fine. To complete the problem you can compute the domain of the inverse. $\endgroup$ – orole Jan 25 '18 at 23:19
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    $\begingroup$ Also it should be $\dots /(2\log(4))$, not $\dots/2\log(4)$. See here for a mathjax guide for correct typesetting. $\endgroup$ – anderstood Jan 25 '18 at 23:22
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    $\begingroup$ Would it kill you to put your code between \$ signs? $\endgroup$ – fleablood Jan 25 '18 at 23:38
  • $\begingroup$ Yes, the computation seems correct. $\endgroup$ – Berci Jan 25 '18 at 23:49

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