Question

using proper mathematical terminology and different reasoning explain why there are no (as I did not learn complex/imaginary solutions yet and the question does not ask for them) real solutions.
a) log_2(-8) = x b) log_-2(8)=x
For a I wrote It has no real solutions as 2 to the power of anything will be positive so 2^x can be equal to -8

For b I can't think of anything that is the mirrored version of what I said for a.

Answer 1

Agree with you a)
To add to this:
y = 2^x and y = log₂ x
are inverse relations of each other.
i) in y = 2^x , y is positive for any value of x > 0
ii) in y = 2^x, y = 1 when x = 0 by definition
iii) if x < 0 , then we have y = 2^-x , we can write it as y = 1/2^x, and
that of course is positive

Therefore y has to be positive

and since y = log₂ x is the inverse of y = 2^x,
x has to be positive

b) consider y = (-2)^x
if x = 1, y = -2
if x = 2 , y = +4
if x = 3, y = -8
if x = 1/2, y = √-2, which is undefined
if x = 1/3 , y = cuberoot(-2) = -1.25992... which is defined
if x = (-2)^2.6 = my calculator says undefined
if x = (-2)^(5/3) = [ (-2)^(1/3)]^5 = -3.1748... , it is REAL

As you can see, our answer bounces all over the place,
if your base is negative and your exponent is "even based" , we get no real
if your base is negative and your exponent is "odd based", we get a real #

Meaning of "odd or even based" :
any fraction a/b can be written as <b> a*1/b</b?
if b is odd, a/b is an odd-based fraction, if b is even .....

since x = (-a)^y <------> y = log_-a x
mathematicians do a cop-out here and defined the base
of a log to be a positive number, x ≠ 1
(what would happen if x = 1 ?? )

What do you think?

Answer 2

"if your base is negative and your exponent is "even based" , we get no real #
if your base is negative and your exponent is "odd based", we get a real #"
Seems to be different enough to work if I reword it to fit the question thanks.

Answer 3

nvm

Answer 4

Solved it