SOLUTION: Find all real values of x such that log base 2 of (log base 2 of x)= log base 4 of (log base 4 of x)

Question 325281: Find all real values of x such that log base 2 of (log base 2 of x)= log base 4 of (log base 4 of x)
Answer by Theo(13342) (Show Source): You can put this solution on YOUR website!
log(2,x) = log(4,x)

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

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.

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