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I have an equation:

$$2^{x - 1} = \frac{360}{y}$$

I want to manipulate it so that $x$ is on the LHS of the equal sign, all by itself. Do you think I remember how to do that?

Any ideas?

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For starters, take the logarithm. – El'endia Starman Aug 26 '11 at 0:16
I wonder how long our "helpful" users can resist going on, without waiting for Peter to return and show his efforts? – GEdgar Aug 26 '11 at 0:25
And you might want to omit all the noob/need to go back to school/shame on me/do you think I remember stuff, nobody cares. – Did Aug 26 '11 at 0:26
@GEdgar, excellent question but we already know the answer: around 12 minutes. – Did Aug 26 '11 at 0:27
And Peter, no, I do not think you remember how to do that. – The Chaz 2.0 Aug 26 '11 at 1:53
up vote 4 down vote accepted

The standard way of dealing "bringing down" variables from exponents is to use logarithms.

The basic properties of the logarithms (either $\log$, the logarithm base 10, or $\ln$, the logarithm base $e$), are:

  • $\ln$ is defined only for positive numbers as inputs.
  • $\ln(ab) = \ln(a)+\ln(b)$;
  • $\ln(\frac{a}{b}) = \ln(a) - \ln(b)$;
  • $\ln(a^b) = b\ln(a)$.

So, if you take logarithms on both sides of your equation, you get $$\begin{align*} 2^{x-1} &= \frac{360}{y}\\ \ln\left(2^{x-1}\right) &= \ln\left(\frac{360}{y}\right)\\ (x-1)\ln(2) &= \ln(360) - \ln(y) \\ x\ln(2) - \ln(2) &= \ln(360) - \ln(y). \end{align*}$$ I trust that at this point you know how to isolate the $x$...

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Fantastic, thank you very much. Is there a reason why you chose ln rather than log base 10? – Peter Sankauskas Aug 26 '11 at 3:27
@Peter: I prefer what is natural to what is common. (-: Logarithm base 2 would be the simplest thing to use here (assuming you have a simple way of computing logarithms base 2), but you can use any base. – Arturo Magidin Aug 26 '11 at 3:38

The first thing you want to do is take the logarithm base two of both sides, which should leave you with $$x-1=\log_2(360/y)$$ Thus, $$x=\log_2(360/y)+1$$

Hopefully this helps, but more generally and for future reference, when $$a^{f(x,y)} = g(x,y),$$ then $$f(x,y)=\log_ag(x,y).$$

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