Santa Barbara Math Tutor

You can download a spreadsheet to accompany this explanation into which you can enter your own values (lines 8 =  10) here

Common wisdom says you can’t use a negative base in a logarithm if you’re working with real numbers

i.e. Y=log_-b(X) is not valid, where –b is a negative number

However, consider this

For every integer a, there is some negative number –p, such that

1) (-p)^(2a+1) = -b : because 2a+1 is an odd number and a negative number to an odd power is a negative number.

P^{(2a + 1)} = b : (the relationship between positive p and positive b)

P = b^1/(2a+1) (spreadsheet line 14 solves for –p)

There is some z, such that

P^z = X  (z solved for on spreadsheet line 16)

If  c = int(z/2) (c is on line 18 of the spreadsheet) then

X = p^(z-2c) * p^2c

Because the exponent of p^2c is even, we can rewrite it as (-p)^2c

2) Let W = (-p)^2c , (spreadsheet line 22) therefore

X = p^(z-2c) * W

we know that (-p)^2c is positive because the exponent 2c is an even integer

Taking log_-b of expression 1) yields

(2a+1) log_-b(-p) = log_-b(-b) : since the log of a number to its base is 1, we have

3) Log_-b(-p) = 1/(2a+1)

Taking log_-b of expression 2 yields

Log_-b(W) = log_-b[(-p)^2c] = 2c * log_-b(-p)

Substituting 1/(2a+1) for log_-b(-p) from equation 3 yields

Log_-b(W) = 2c/(2a+1)

As a approaches infinity, p approaches one, p² approaches 1, W approaches X and log_-b(W) approaches log_-b(X), so

Log_-b(X) = 1/(2a+1) * 2c

4) Log_-b(X) = 1/(2a+1) * [2 * int( log_b(X) * (2a+1)/2)]

You can see this formula implemented step by step in the spreadsheet starting on line 27

The answer can be checked (starting on spreadsheet line 37) by the following process

Log_-2(X) = 1/(2a+1) * 2c is

X =((-b) ^{1/(2a + 1)}) ^2c

First, take the (2a + 1) root of –b which is the same as minus the (2a+1) root of b.

Then take that number to the 2c power.

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