We will show that the logarithm of a negative number or zero is undefined or does not exist.
Similarly, you cannot find the logarithm of the following expressions.
log_{5} -125
log_{10} -100
log_{5} 0
log_{4} 0
The reason for this is that any positive number b raised to any power x cannot equal to a number y less than or equal to zero.
The definition says If y = b^{x}, then log_{b} y = x
If x is bigger than zero or x is equal to zero, It is obvious that b^{x} will be bigger than zero as you can see in the examples below. As a result, y will also be bigger than zero since y = b^{x}
65^{0} = 1
7^{0} = 1
24^{1} = 24
2^{3} = 8
4^{2} = 16
5^{3} = 125
6^{4} =1296
8^{5} = 32768
How about when x is negative?
Let x = -2, -3, and -8 and let b = 5
y = 5^{-2} = 1 / 5^{2} = 1 / 25 = 0.04
y = 5^{-3} = 1 / 5^{3} = 1 / 125 = 0.008
y = 5^{-8} = 1 / 5^{8} = 1 / 390625 = 0.00000256
As you can see, although y can get very small or very close to zero, it will never be equal to zero or worse be a negative number. That is the key concept here!
Since y can never be zero or negative, it does not make sense to replace y in
log_{b} y with zero or a negative number.
Now, you can clearly see why these expressions do not make sense
log_{5} -125 log_{10} -100 log_{5} 0 log_{4} 0
In fact, for log_{5} -125, there is no number x, such 5^{x} = -125
If you choose 3, you will get 5^{3} = 125 and if you choose -3, you will get 5^{-3} = 0.008
If it did not work for x = 3 and x = -3, no other numbers will work!
By the same token, for log_{5} 0, there is no number x such that 5^{x} = 0
Jan 26, 23 11:44 AM
Jan 25, 23 05:54 AM