Why can log3(-9) not be evaluated????

Logarithms of negative numbers do not exist in the real domain. If you are familiar with complex numbers and their properties you CAN evaluate this in the complex domain.

log3(-9) would yield the power of three that gives -9

i.e. 3^x = -9

there is no real number x, that will give the desired result. In fact there is no real value of x that will result in a negative number.

because negative log numbers are not possible.

logx only defined for x>0

log_3(-9) = x
-9 = 3^x

If x = 2:
3^x = -9
3^2 = -9
9 ≠ -9

If x = -2
3^x = -9
3^-2 = -9
1/3^2 = -9
1/9 ≠ -9

Because the logarithms of negative numbers do not exist.
There is no power to which 3 could be raised to make -9

Negative log numbers are not possible. To give an example using your value, lets say
log3 (-9) = x
using log properties,
-9 = 3^x
If you look in the above equation, whatever x is, 3 to the power x cannot turn out to be a negative value. If x is positive, a positive value will result. If x is negative, a decimal positive value will result. -9 never equals 3^x. That's why log3(-9) can't be evaluated. Negative log values are not defined.

suppose that log[3](-9) = something, like x. Then according to properties of logarithms relating logarithms to exponentials, we have that

log[3](-9) = x implies that 3^x = -9. This means that 3 raised to a power is going to give you a negative number. If you try plugging in a few numbers into "3^x", both positive and negative, you'll find that you'll never get negative numbers. If you graph "3^x" on a graphing calculator or by hand, you'll see that the graph is always above the x-axis, which means that the outputs are always positive, so "3^x" can never be positive for any "real" x. Somebody on here said something about complex numbers. Turns out that in the system of complex numbers, it is possible to find an x for which "3^x" is negative. Don't worry about that now ;)...