Certainly! Let's proceed step-by-step to rewrite the given equation in logarithmic form.
1. We start with the given equation:
[tex]\[
7^{-1} = \frac{1}{7}
\][/tex]
2. To rewrite this equation in logarithmic form, we need to understand the relationship between exponents and logarithms. The logarithmic form [tex]\(\log_b(a) = c\)[/tex] is equivalent to the exponential form [tex]\(b^c = a\)[/tex] where [tex]\(b\)[/tex] is the base, [tex]\(a\)[/tex] is the result, and [tex]\(c\)[/tex] is the exponent.
3. In our given equation, the base [tex]\(b\)[/tex] is [tex]\(7\)[/tex], the result [tex]\(a\)[/tex] is [tex]\(\frac{1}{7}\)[/tex], and the exponent [tex]\(c\)[/tex] is [tex]\(-1\)[/tex].
4. Using the logarithmic form [tex]\(\log_b(a) = c\)[/tex], we can rewrite the given equation as:
[tex]\[
\log_7\left(\frac{1}{7}\right) = -1
\][/tex]
Therefore, the rewritten logarithmic equation is:
[tex]\[
\log_7\left(\frac{1}{7}\right) = -1
\][/tex]
This shows that the logarithm base 7 of [tex]\(\frac{1}{7}\)[/tex] is [tex]\(-1\)[/tex].
