Question 1081007
using the desmos.com calculator, you can graph this equation as follows:


y = log2(x)


y = log(x) / log(2)


x = 2^(-y)


all 3 equations will create the identical graph.


this is because:


the log conversion formula says that log2(x) = log(x) / log(2)


the basic definition of logs says that y = log2(x) if and only if 2^y = x


your equation is y = -log2(x)


the base conversion formula becomes y = -log(x) / log(2).


the basic definition of logs says that y = log2(x) if and only if 2^y = x.


but your equation is y = -log2(x)


multiply both sides of this equation by -1 and you get:


-y = log2(x)


now use the basic definition to get -y = log2(x) if and only if 2^(-y) = x


this is the same as x = 2^-y which was graphed.


the alternative way to derive this is a little more convoluted but gets you the same equivalence.


start with y = -log2(x)


by the rules of logs, this is equivalent to y = -1 * log2(x) which is the same as y = log2(x^-1)


by the basic definition of logs, this is true if and only if 2^y = x^-1


since x^-1 is the same as 1/x, this becomes 2^y = 1/x


since 2^y is the same as 1/(2^-y), this becomes 1/(2^-y) = 1/x


this is true if and only if 2^-y = x which is the same as x = 2^-y


the graph is shown below:


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