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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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    All I hear is "blah blah blah I don't have a personality blah blah blah". As you can see below, some people here are useful - you should take note.2011-08-26

2 Answers 2

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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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    @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.2011-08-26
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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 $f(x,y)=\log_ag(x,y).$