Question
Why is any logarithmic expression with an argument of 1 equivalent to 0? Writen like 8 to h g5ad Dr r
Answers
GPT-4o mini
The reason any logarithmic expression with an argument of 1 is equivalent to 0 can be explained using the definition of logarithms.
The logarithm of a number \( b \) to a base \( a \) (denoted as \( \log_a(b) \)) answers the question: "To what power must the base \( a \) be raised to obtain the number \( b \)?"
Mathematically, this is expressed as:
\[
\log_a(b) = c \iff a^c = b
\]
When the argument \( b \) is 1, we can express this as:
\[
\log_a(1) = c \iff a^c = 1
\]
For any positive base \( a \) (where \( a > 0 \) and \( a \neq 1 \)), the only exponent that satisfies the equation \( a^c = 1 \) is \( c = 0 \), because:
\[
a^0 = 1
\]
Therefore, regardless of the base \( a \), as long as it is a valid logarithm (positive and not equal to 1), we have:
\[
\log_a(1) = 0
\]
This is true for all logarithmic expressions with an argument of 1.
The logarithm of a number \( b \) to a base \( a \) (denoted as \( \log_a(b) \)) answers the question: "To what power must the base \( a \) be raised to obtain the number \( b \)?"
Mathematically, this is expressed as:
\[
\log_a(b) = c \iff a^c = b
\]
When the argument \( b \) is 1, we can express this as:
\[
\log_a(1) = c \iff a^c = 1
\]
For any positive base \( a \) (where \( a > 0 \) and \( a \neq 1 \)), the only exponent that satisfies the equation \( a^c = 1 \) is \( c = 0 \), because:
\[
a^0 = 1
\]
Therefore, regardless of the base \( a \), as long as it is a valid logarithm (positive and not equal to 1), we have:
\[
\log_a(1) = 0
\]
This is true for all logarithmic expressions with an argument of 1.