Question 325281
log(2,x) = log(4,x)


If we graph these equation, we can see that they are equal when x = 1.


{{{graph(600,600,-10,10,-10,10,log(2,x),log(4,x))}}}


When x = 1, log(2,x) = log(2,1) = log(10,1)/log(10,2) = 0


When x = 1, log(4,x) = log(4,1) = log(10,1)/log(10,4) = 0


That's the only number where they are equal.


Per the graph it looks like they would also be equal when x = 0, but the log(x) when x = 0 is not valid.


log(2,0) = log(10,0)/log(10,2) = Error - the calculator won't compute it.


If you use the exponential definition of a log, then y = log(b,x) if and only if b^y = x


If x = 0, this means that b^y = 0


Since there is no value of y such that b^y = 0, then you can't get the log of it.


Example:


Let b = 2


2^y = 0


If y = 0, then 2^y = 2^0 = 1


If y is any other positive number, then 2^y <> 0.


If y is any other negative number, then 2^-y <> 0.


It can approach 0 but it can't be equal to 0.


Example:


2^.00000000000000000000001 will be something greater than 1 since 2^0 = 1.


2^-99999999999 = 1/2^99999999999 = something smaller than 1 but greater than 0 since 1 divided by any number will always be greater than 0.


Bottom Line is that log(2,x) = log(4,x) when x = 1.