Logarithm of 1

For any base b (as long as b is greater than 0 and not equal to 1), the logarithm of 1 is always 0.

That’s because any number raised to the power of 0 equals 1.

We can write this as:

\log_b(1) = 0 \quad \text{for any } b > 0 \text{ and } b \neq 1

\log_{-4}(1) is undefined because we can’t find the logarithm of a negative base.
Answer: undefined

Natural logarithms of 1

Because we can rewrite \ln(1) as \log_e(1), we can evaluate it using the same principle.

We know that \log_e(1) = 0 because e^0 = 1.

So \ln(1) = 0.

Question	Answer
What is the value of \log_b(1) for any valid base b?	\log_b(1) = 0 for any b > 0 and b \neq 1, because b^0 = 1.
Evaluate \log_2(1).	\log_2(1) = 0 because 2^0 = 1.
Evaluate \log_{10}(1).	\log_{10}(1) = 0 because 10^0 = 1.
Evaluate \log_e(1).	\log_e(1) = 0 because e^0 = 1.
Evaluate \log_{0.5}(1).	\log_{0.5}(1) = 0 because (0.5)^0 = 1.
Evaluate \log_{-4}(1).	\log_{-4}(1) is undefined because the base is negative.
Find the value of x such that \log_3(x) = 0.	x = 3^0 = 1.
What is the value of \ln(1)?	\ln(1) = 0, because \ln(1) = \log_e(1) and e^0 = 1.
Solve for x: \ln(x) = 0.	x = e^0 = 1.