Explore the limitations of the values of b and x in equations of the form logbx=L

by determining which logarithm is undefined.(1 point)
Responses

log1319
logarithm subscript Start Fraction1 over 3 End Fraction baseline Start Fraction 1 over 9 End Fraction

log51
logarithm subscript 5 baseline 1

log2.56.25
logarithm subscript 2.5 baseline 6.25

log50

1 answer

To determine which logarithm is undefined among the given options, we can analyze the values of \( b \) (the base) and \( x \) (the argument of the logarithm) for each logarithmic expression according to the rules for logarithms:

  1. The base \( b \) must be greater than 0 and cannot be equal to 1 (i.e., \( b > 0 \) and \( b \neq 1 \)).
  2. The argument \( x \) must be greater than 0 (i.e., \( x > 0 \)).

Let’s evaluate each option:

  1. \( \log_{13} 19 \): Here, \( b = 13 \) (valid) and \( x = 19 \) (valid). This expression is defined.

  2. \( \log_{\frac{1}{3}} \frac{1}{9} \): The base \( b = \frac{1}{3} \) (valid since \( b > 0 \) and \( b \neq 1 \)), and the argument \( x = \frac{1}{9} \) (valid since \( x > 0 \)). This expression is defined.

  3. \( \log_{5} 1 \): The base \( b = 5 \) (valid), and the argument \( x = 1 \) (valid since \( x > 0 \)). This expression is defined.

  4. \( \log_{2.5} 6.25 \): The base \( b = 2.5 \) (valid), and the argument \( x = 6.25 \) (valid since \( x > 0 \)). This expression is defined.

  5. \( \log_{5} 0 \): The base \( b = 5 \) (valid), but the argument \( x = 0 \) (invalid, as \( x > 0 \) is required). This expression is undefined.

Based on this analysis, the logarithm that is undefined is:

\( \log_{5} 0 \).