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.
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 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?
"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.
nvm
Solved it
To explain why there are no real solutions for the equations a) log_2(-8) = x and b) log_-2(8) = x, we need to understand the properties of logarithmic functions and the concept of domain and range.
a) In the equation log_2(-8) = x, the base of the logarithm is 2. Logarithms require both the base and the argument (the number inside the logarithm) to be positive. This is because logarithms are the inverse function of exponentiation, and raising positive numbers to any power will always result in a positive number.
Therefore, in this case, we have a negative argument (-8) for a logarithm with a positive base (2). Based on the above property, there is no real number x that can be the solution for this equation. This is because 2 raised to any power will always be positive, so there is no value of x that makes 2^x equal to -8.
b) In the equation log_-2(8) = x, we have a negative base (-2). Similarly to the previous case, the base and the argument must both be positive. However, logarithms with negative bases are only defined when the argument is negative.
In this case, the argument (8) is positive, which means it is not valid to take the logarithm of 8 with a negative base (-2). Therefore, there are no real solutions for this equation either.
In summary, both equations a) and b) have no real solutions because the properties of logarithmic functions do not allow for negative bases or positive arguments.