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?
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?
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:
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 $x$...
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 $$f(x,y)=\log_ag(x,y).$$