Given equation: $T_n = t_1 \cdot q^{n-1}$

The equation goes like this (some random values I put):

$$435 \cdot 2^{n-1}$$

How can you find the $n$ number if it's squared (or whatever you call this in english)? I have tried many ways, but no luck, it's not like finding a simple X variable.

Note, this is not homework, but for myself to understand.

If anyone can explain it, thanks!

  • $\begingroup$ do you mean $2^n-1$ or $2^{n-1}$? $\endgroup$ – Thanos Darkadakis Mar 25 '14 at 12:16
  • $\begingroup$ Your $\;T_n=T_1q^{n-1}\;$ is the equation for the general term in a geometric sequence. The rest of your question is not comprehensible to me. $\endgroup$ – DonAntonio Mar 25 '14 at 12:16
  • $\begingroup$ @naslundx, how do you know the OP meant $\;2^n-1\;$ and not $\;2^{n-1}\;$ ?? Lucky guess or what? $\endgroup$ – DonAntonio Mar 25 '14 at 12:18
  • $\begingroup$ @DonAntonio it's the second one, n - 1 $\endgroup$ – Jonn Mar 25 '14 at 12:19
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    $\begingroup$ Then you better render nothing if you're not sure, @naslundx, and wait until the OP addresses the doubt. Anyone with some time in this site knows that newbies can be extremely sloppy and careless about the way they write mathematics, and your editing the question may confuse others. $\endgroup$ – DonAntonio Mar 25 '14 at 12:21

If by any luck you meant to find the value of $\;n\;$ in

$$T=T_1q^{n-1}\implies q^{n-1}=\frac T{T_1}\implies (n-1)\log q=\log\frac T{T_1}\implies$$

$$n=\frac{\log\frac T{T_1}}{\log q}+1$$

  • $\begingroup$ What is the value of T? Would be very nice if you can initialize in when lets say T1 = 377 and q = 6, So i have an example in numbers $\endgroup$ – Jonn Mar 25 '14 at 12:29
  • $\begingroup$ @Jonn you want to find the value of $T$? And you know $T_1$ and $q$? $\endgroup$ – Guy Mar 25 '14 at 12:32
  • $\begingroup$ I know T, but I dont understand the differnece between T and T1 $\endgroup$ – Jonn Mar 25 '14 at 12:34
  • $\begingroup$ In my post's case, T1 = 435 and q = 2. so in his example q^n-1 = 435/435 whcih is (n -1) log q? $\endgroup$ – Jonn Mar 25 '14 at 12:35
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    $\begingroup$ Simple substitution, @Jonn: $$T_1=435\,,\,q=2\implies n=\frac{\log \frac T{435}}{\log 2}+1$$ and $\;T\;$ is whatever you want this to be equal to: $$T=435\cdot 2^{n-1}$$ $\endgroup$ – DonAntonio Mar 25 '14 at 12:44

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