Rewrite 36=729 as a logarithmic equation.(1 point) Responses log7296=3 log subscript 729 baseline 6 equals 3 log36=729 log subscript 3 baseline 6 equals 729 log6729=3 log subscript 6 baseline 729 equals 3 log3729=6 log subscript 3 baseline 729 equals 6

1 answer

To rewrite the equation \( 36 = 729 \) as a logarithmic equation, we can express this in the logarithmic form.

Recall that the general form of converting an exponential equation \( a = b^c \) into a logarithmic form is given by:

\[ c = \log_b(a) \]

Since \( 36 \) cannot be expressed as \( 729^c \) for a straightforward integer \( c \) directly, we can analyze the original equation in other terms.

However, if your intention is to express a relationship between \( 729 \) and some power that results in \( 36 \), we should observe:

If \( 6^2 = 36 \) and \( 729 = 3^6 \), the logarithmic formulation becomes relevant in deducing powers.

Let’s derive this correctly:

  1. Recognize \( 36 \) and \( 729 \) in terms of their bases.
  2. \( 6^2 = 36 \) can be expressed as a logarithm.

So, we can state:

\[ \log_{729}(36) = \frac{2}{6} = \frac{1}{3} \]

Thus, the correct logarithmic form based on the different values yields:

\[ \log_{729}(36) = \frac{1}{3} \]

If options were provided directly, based on standard conventions in logarithms, this can yield variations leading to explicit forms involving base transformations.

Given the patterns of the responses available, one plausible logarithmic relationship is:

Option: \( \log_{729}(36) = \frac{2}{6} = \frac{1}{3} \)

This rephrased approach yields: \[ \log_{729}(36) = \frac{1}{3} \]

If a consideration is purely on existent responses, ensure to select a representation independently satisfying the equivalency derived. If you still need clarity on another option, please provide full context.

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