Log_{-2} 8 has no solution?

Question

or no real solution? I get a mental block when I stare at (-2)^x = 8, please help.

Answer 1

No real solution.

The equation x = log_{-2} 8 does have a solution in the complex numbers. (If you've taken a course in complex analysis, the below is how to find it. You can ignore the work below if you haven't taken a course in complex analysis.)

log_{-2} (8) = x

(-2)^x = 8

x ln (-2) = ln 8

x = (ln 8) / ln (-2)

x = (ln 8) / (ln 2 + i Arg (-2))

x = (ln 8) / (ln 2 + pi * i).

Answer 2

No real solution

Answer 3

log -2 = x is 10^x = -2 which isn't true for any x (where x is any real number)

Answer 4

You must first know how logarithms are defined. It goes like this. If a > 0 and a not equal to 1 and x is any real number and y = a^x then the logarithm function is defined as x = log(base a) y. From this definition it is clear that logarithms are not defined for negative bases. So you can say now your problem has no solution in the set of real numbers. That is to say there is no real solution. Remember while a > 0 and a not equal to 1 and x is any real number a^x is always positive. That is y > 0. Thus logarithms are not defined for negative numbers even though the base may be positive.